Problems-Bodies in vertical motion.

1.A stone is projected vertically upwards with a velocity 29.4m/sec. i)Calculate the maximum height to which it rises  ii) calculate the time taken to reach maximum height iii) Find the ratio of velocities after 1sec, 2 sec, 3sec of its journey.

Soln: From the data given in the problem,

Initial velocity of stone u=29.4 m/sec,

Acceleration due to gravity g=9.8 m/sec^2,

i) H_{max} = \frac{u^2}{2g} = \frac{(29.4)^2}{2(9.8)}

H_{max} = \frac{29.4}{2} = 14.7 m.

ii) Time of ascent T_a =\frac{u}{g} = \frac{29.4}{9.8} = 3sec.

iii) Ratio of velocities of a vertically projected  up body after 1 sec, 2 sec, 3sec , . . . . . . . . . . . . . . of its journey will be v_1:v_2:v_3 . . . . . . . . . . . . :v_n = (u-g) : (u-2g) : (u-3g) : . . . . . . . . : (u-ng).

v_1:v_2:v_3 = (29.4-9.8 ) :  (29.4- 19.6) :  (29.4 – 29.4)

v_1:v_2:v_3= 19.6 : 9.8 : 0 .

2.A stone is dropped from  the top of a tower.The stone touches the ground after 5sec.Calculate the i)height of the tower and the  ii) velocity with which it strikes the ground.

Soln: From the data given in the problem,

Initial velocity of the stone  u = 0 m/sec,

Acceleration due to gravity g = 9.8$latex  m/sec^2$

Time of descent t_a = 5 sec.

i) Height of the tower be S=H (say)

substitute the above values in the equation  S=ut+ \frac{1}{2}at^2

H = 0(5)+\frac{1}{2}(9.8)(5)^2 = 0 + \frac{1}{2}(9.8)(25) = (4.9)(25) = 122.5 m.

ii) Let the velocity when it touches = v (say),

substitute the above values in the equation v=u +gt_a

v = 0+(9.8)(5) = 49 m/sec.

3.A stone is projected vertically upwards with an initial velocity 98 m/sec . Calculate i) maximum height the body reaches ii ) find its velocity when it is exactly at the mid point of it’s journey iii) time of flight.

Soln : From the data given in the problem,

Initial velocity of the stone u = 98 m/sec,

acceleration due to gravity a = 9.8 m/sec^2,

i) Maximum height H_{max} = \frac{u^2}{2g} = \frac{(98)^2}{2(9.8)} =  490 m.

ii) When it is in the middle S= H_{max}{2} =245 m

Let the velocity = v (say),

substitute the values in the equation v^2 - u^2 = 2gS,

we get v^2 - (98)^2 = 2(-9.8)(245),

v^2 = 9604 – 4802 =4802 ,

v =69.30 m/sec

iii) time of flight T = \frac{2u}{g} = \frac{2(98)}{(9.8)} = 20 sec.

4.Estimate the following.
(a) how long it took King Kong to fall straight down from the top of a 360 m high building?_________ seconds
(b) his velocity just before “landing”___________ m/s.
( Question by ELISE in Yahoo answers).

Soln: From the data given in the problem,

Height of building  S=H = 360m,

Acceleration Due to gravity g=10 m/sec^2,

i) time of descent be t_d= ?

substitute the values in the formula   S=ut+\frac{1}{2}gt^2

we get  360 = (0)(t_d) +\frac{1}{2}(10)(t_d)^2 ,

360 = 5 t_d^2 ; t_d^2 = 360/5=72

Therefore t_d = 8.49  sec.

ii )Velocity of king kong just before touching the ground v =gt_d,

v = (10)(8.49) =84.9 m/sec.

5. A baseball is hit  straight up into the air with a speed of 23 m/s.
(a) How high does it go?____ m
(b) How long is it in the air?_____ s.
( Question by ELISE in Yahoo answers).

Soln: From the data given in the problem,

Acceleration Due to gravity g=10 m/sec^2,

a) Maximum height H_{max}= \frac{u^2}{2g}\frac{(23)^2}{20}

H_{max}= 529/20 = 26.45 m.

b ) Time of flight of base ball T = \frac{2u}{g},

T= \frac{(2)(23)}{10}, = 46/10 = 4.6 sec.

6.A body  starting from rest slides down an inclined plane.Find the velocity after it has descended vertically a distance of 5metres. (g=9.8 m/sec^2).

Soln: The velocity of the body when it touches the ground sliding down an inclined plane ,will be same as when the body vertically falls freely from height ‘h”.

From the problem height S=h=5m; acceleration due to gravity g=9.8m/sec^2; Initial velocity u=0 ; V=?

substitute the values in the equation V^2-u^2 = 2gs ,

we get V^2-0^2 = 2(9.80) (5) ; V^2 = 98

V =9.899 m/sec.

7.A ball projected vertically upwards returns to ground after 15sec.Calculate i)maximum height to which it rises ii)velocity with which it is projected and iii)its position after 6 seconds.

Soln:From the data given in the problem,

Velocity of projection u=?

acceleration due to gravity g=9.8m/sec^2,

time of flight T =15 sec

substitute the values  of T,g in the equation T = \frac{2u}{g}

\frac{2u}{9.8} =15 ; 2u = 15 (9.80) = 147

ii )Velocity with which it is projected   u= 73.5 m/sec.

i)Maximum height H_{max} =\frac{u^2}{2g},

substitute the values of  u,g in the equation we get   H_{max} = \frac{73.5^2}{2(9.8)}

H_{max} = \frac{5402.25}{19.6} =275.625 m

1iii) Let Its  position after   t=6sec be  S

substitute the values of u,t and g in the equation S=ut – \frac{1}{2} at^2

S = (73.5)(6) – \frac{1}{2} 9.8(6)^2=441-176.4 = 264.6 m high from the ground.

8.A stone was dropped from a rising baloon at a height of 150m above the ground and it reaches the ground in 15 seconds.Find the velocity of the baloon at  the instant the stone was dropped.

Soln: From the data given in the problem,

Acceleration due to gravity g=9.8 m/sec^2,

Height of the balloon when the stone is dropped from it  h=150m,

Time of flight T=15 sec,

Let the velocity of the balloon when the stone is dropped from it is  u (say).

Substitute the values of g,t and h in the equation  h = \frac{1}{2} gt^2 -ut

we get 150 = \frac{1}{2} (9.8)(15)^2-u(15) = (4.9)(225) – 15u,

150 = 15[(4.9)(15)- u],

10=73.5 – u ; u = 73.5- 10 =63.5 m/sec.

9. From the top of a tower of 200 metres high, a stone is projected vertically upwards with a velocity 49m/sec.Calculate the i)maximum height traveled by it from ground level, ii)the velocity with which it strikes the ground and the iii) time it takes to reach the ground.

Soln: From the data given in the problem,Height of the tower is h=200m, picture

velocity of projection  u = 49 m/sec,

Max height of the stone from top of the tower be S= X (say),

velocity at maximum height v = 0,

i) substitute the values of u,v and g in equation V^2-u^2 = -2gs,

we get 0^2-49^2 = -2(9.8)X,

X = \frac{2401}{19.6} = 122.5 m from the top of the tower

Maximum height from ground level H = X +h =122.5 + 200 = 322.5 m.

ii) Let the velocity with which it reaches the ground be v_1 say,

substitute the values in the equation v_1= \sqrt{2gH}

v_1= \sqrt{2(9.8)(322.5)},

v_1= \sqrt{6321},

v_1= 79.50 m/sec.

iii) Time of flight T =?

Substitute the values  u,g and h in the equation     h = \frac{1}{2} gt^2 -ut,

we get 200 = \frac{1}{2} (9.8)T^2 -49T,

200 = (4.9)T^2 – 49T

T^2– 10T = \frac{200}{4.9},

by solving this quadratic equation for T we get T= 13.11 sec


Problems Linear motion.

In this post and  in few of my posts to come, I would like to solve problems on linear motion,freely falling bodies,vertically projected up bodies and projectiles .

1.An object accelerates from rest to a velocity 20m/sec in 4seconds.If  the   object  has uniform acceleration, find its acceleration and displacement in this time.

Soln: From the data given in the problem we have,

Initial velocity = u =0,

final velocity v=20 m/sec,

Time of journey t=4sec,

Acceleration a = \frac{(v-u)}{t} = \frac{(20-0)}{4}= 5 msec^{-2}

Displacement S = ut + \frac{1}{2} at^2 = 0 (4)+\frac{1}{2}\times5\times4^2

S = 40m.

2.An object starting from rest moves with uniform acceleration of 3msec^{-2} for 6sec.Find its velocity and displacement after 6seconds.

Soln: From the data given in the problem we have,

Initial velocity of the object u = 0,

Acceleration of the object a= 3msec^{-2},

Time of journey t =6sec.

Final velocity of the object v=u+at =0 + 3(6) = 18 m/sec.

Displacement S= ut + \frac{1}{2} at^2 = 0 (6)+\frac{1}{2}\times3\times6^2

S= 54m.

3.An object starting from rest moves with uniform acceleration of 4msec^{-2}.Find its displacement i) 5seconds   ii) in 5th second iii) 8th second.

Soln : From data given in the problem

Initial velocity of the object u=0,

Acceleration of the object a=4msec^{-2},

i) time t=5 seconds,

Displacement of the object S=ut + \frac{1}{2} at^2 = 0 (5)+\frac{1}{2}\times4\times5^2

Displacement of the object in 5seconds  S= 50m.

ii) Displacement of the body in 5th second =?

Let us substitute n=5 in the formula S_n=u+a(n-1/2)

S_5=0+4(5-1/2) = 4(4.5) =18m.

iii)Displacement of the body in 8th second =?

Let us substitute n=8 in the formula S_n=u+a(n-1/2)

S_8=0+4(8-1/2) = 4(7.5) =30m.

4.An object started moving with an initial velocity of 10m/sec, after traveling a distance of 5m  gets a velocity 20m/sec.Find its i) acceleration ii) time taken for 5m displacement.

soln: From the data given in the problem,

Initial velocity of the object u=10m/sec,

Final velocity of the object    v= 20m/sec,

Displacement S=5m,

i) acceleration a=?

Substitute the values of u,v and S in the equation V^2-U^2 =2as,

we get (20)^2-(10)^2=2a(5)

300 = 10a  or a = 300/10=30msec^{-2}.

ii)Time t=?

Substitute the values of u,v and a in the equation v=u+at,

we get  20=10+(30)t ;  10=30t

t =10/30 = 1/3 = 0.333 sec.

5.When an observer started observing a car it’s velocity was x m/sec , if it travels for 10sec with uniform acceleration 2.5msec^{-2} and its velocity increases to 75m/sec.Find i) Initial velocity of the car ii) Displacement of the car in 10sec iii) Displacement of the car in first 5sec and last 5sec, what is your inference.

Soln: From the data given in the problem,

Initial velocity of the car u = x (say),

Final velocity = 75m/sec,

Acceleration a = 2.5msec^{-2},

Time of journey t=10sec.

i) Substitute the values of u,v,a and t in the equation v=u+at

we get   75 = x+2.5(10) ; x=50 m/sec.

ii) Let the displacement of the car in 10sec  be S

Substitute the values of u,a and t in the equation S=ut +\frac{1}{2} at^2

We get    S = 50(10)+\frac{1}{2} (2.5)(10)^2 ;

S = 500+125 ; s=625 m

ii) Let the displacement in first 5sec be S_1

Substitute t=5 sec and thve values of u and a in the equation S_1=ut +\frac{1}{2} at^2

we get S_1=50(5) +\frac{1}{2} (2.5)(5)^2

S_1 = 250 + 31.25 = 281.25 m.

Let the displacement in  next  5sec be S_2

S_2 = S –S_1

S_2 = 675 – 281.25 = 393.75 m

We can observe that, even though the time of journey is same , S_2>S_1

Displacement of the body in second half  S_2 is greater than in the first half time S_1  of its journey.

6 .A cheetah can accelerate from rest to 24.0 m/s in 6.70 s.

Assuming constant acceleration, how far has the cheetah run in this time?(Question by suzi in yahoo answers).

Soln: From the data given in the problem,

Initial velocity of the cheetah  u=0,

Final velocity of cheetah   v=24 m/sec,

Time t = 6.7sec

Distance traveled by cheetah  be S=?

The equations of motion are v^2-u^2 = 2aS- – – – –   (1)

and  a = \frac{(v-u)}{t}  – – –  –  –  (2)

From equations (1) and (2) we get v^2-u^2=2\frac{(v-u)}{t}S

simplifying it we get   S = \frac{(v^2-u^2)}{2}\frac{t}{(v-u)} = \frac{(v+u)}{2}\times t

Substitute the values of u,v and t    we get S = \frac{(24+0)}{2}\times6.7 =12(6.7)=80.4 m

7) A car covers first halt of the distance between two places at a speed of 30 km/hr and the second half at a 90km/hr.What will be the average speed of the car?

Soln: Method I: Let the total distance between the places be S.

Time taken to cover First half  t_1= \frac{S}{2\times30} = \frac{S}{60} hours.

Time taken to cover Second  half  t_1= \frac{S}{2\times90} = \frac{S}{180} hours.

Total time   t = t_1+ t_2 =\frac{S}{60}+\frac{S}{180}=\frac{4S}{180} = \frac{S}{45}.

Average speed = Total distance /total time.

= S/t = \frac{(S)(45)}{S}

= 45km/hr.

Method II (Short cut):  If the body covers 1st half of distance with a speed x and the second half with a speed y,then the average speed = \frac{2xy}{x+y}.

Average speed  = \frac{2(30)(90)}{30+90} = latex \frac{2700}{60}$ = 45km/hr.

8)  A body starts from rest and acqires a velocity of 400m/sec  in 10seconds.Calculate the acceleration and distance traveled.

Soln: From the data given in the problem

Initial velocity of the body u=0,

Time of journey   t=10sec,

Final velocity v=400m/sec,

Acceleration  a=?  and distance traveled in 10sec   s=?

Substitute the values of u,v and t in the equation   v=u+at,

we get     400= (0)(10) + a (10) ;  10a=400

a= 40m/sec^2

Substitute the  values of u,a and t in the equation S= ut +\frac{1}{2} at^2

S=(0)(10)+\frac{1}{2} (40)(10)^2 = 0+2000 = 2000m.

9) A body moving with uniform acceleration covers 6m in 2^{nd} second and 16m in 4^{th} second.Calculate the initial velocity,acceleration and distance moved in  6^{ th } second .

Soln: From the data given in the problem

Distance moved in 2^{nd} second s_2 = 6m,

Distance moved in 4^{th} second s_4 = 16m,

s_4s_2 = a(4-2) =2a

therefore 2a = s_4s_2=16-6 =10

a = 5 m/s^2

Substitute the values of  s_2,a  and n=2 in the equation s_2 =u+a(n -1/2)

we get    6=u+5(2-1/2) ; 6=u+7.5

u=-7.5 +6 = -1.5 m/sec.

Distance moved in the 6^{th} second = u+a(6-1/2),

Substitute the values of   u,a  in the above equation we get 6^{th} second=-1.5+5(5.5)

6^{th} second =-1.5+27.5 =26m.

10) An object started traveling with a velocity 2m/sec moves with an uniform acceleration of 3 m/s^2.

i) Find the ratio of displacements in   a) 1^{st},3^{rd},5^{th}  seconds  b) 2^{nd},4^{th},  and 6^{th}  seconds .

ii)Find the ratio of velocities   a) 1^{st},3^{rd},5^{th}  seconds  b) 2^{2d},^{th},6^{th}  seconds .

Soln: From the data given in the problem

Initial velocity  u=2 m/sec,

Acceleration   a =3 m/sec^2,

i- a) The ratio of displacements in   1^{st},3^{rd},5^{th}  seconds

From the formula s_1:s_3:s_5 = (2u+a) : (2u+5a) : (2u+9a)

s_1:s_3:s_5 = (4+3) : (4+15) : (4+27)

s_1:s_3:s_5 = 7:19:31 .

i-b) The ratio of displacements in   2^{nd},4^{th},6^{th}  seconds

From the formula s_2:s_4:s_6 = (2u+3a) : (2u+7a) : (2u+11a)

s_2:s_4:s_6 = (4+9) : (4+21) : (4+33)

s_2:s_4:s_6 = 13 : 25 : 37 .

ii -a) From the formula v_1:v_2:v_3: .  .  .  .  .  .  .  . :v_n= (u+a) : (u+2a) : (u+3a) : .   .   .  .  .  .  .  .  . : (u+na).

The ratio of velocities   a) 1^{st},3^{rd},5^{th}  seconds

v_1:v_3:v_5 = (u+a) : (u+3a) : (u+5a) = 5 :11 : 17 .

ii- b) The ratio of velocities   a) 2^{nd},4^{th},6^{th}  seconds

v_2:v_4:v_6 = (u+2a) : (u+4a) : (u+6a) =8 :1 4 :20 .

11) A body travels 200cm in the first two seconds and 220cm in the next four seconds.What will be the velocity at the end of the seventh second from the start?

Soln: The displacement of the body in first 2 sec S_1 =200cm,

Let the initial velocity = u(say) ,

Acceleration = a(say), time t_1 =2sec

S_1 = ut_1+\frac{1}{2} at_1^2,

Substitute the value of  t_1  in above equation, we get

S_1 = 2u+\frac{1}{2} a(2)^2,

S_1 = 2u +2a ; 2(u+a) =200

Therefore          u+a = 100 – – – – – – – – – – – – –  – – – – –  – (1)

Given that the body travels 220cm in next 4sec.That is from the start it displaces  200+220 = 420 cm in  6sec.

Displacement S_2 =420 cm,

time t_2=6 sec,

substitute theses values in the equation S_2 = ut_2+\frac{1}{2} at_1^2,

S_2 = 6u+\frac{1}{2} a(6)^2=6u+18a,

6(u+3a) = 420 ; u+3a = 70 – – – – – – – – – – – – –  – – – – –  – (2)

Solving equations (1) and (2)     or Eq (2) – Eq(1)

we get  2a= -30 ;     a=-15 cm/sec^2,

Substitute value of  a in Eq(1) we get  u-15 = 100,

u = 115 cm/sec.

The velocity at the end of 6{th} second  v=u+at_2

we get    v= 115+(-15)(6) =115-90 =25cm/sec.

Therefore  final velocity v=25cm/sec.

12.A subway train starts from rest at a station and accelerates at a rate of 16.5m/sec^2  for  13.1 sec.It runs at constant speed for 69.7s and slows down at a rate of 3.45 m/sec^2until it stops at the next station.What is the total distance covered?

Soln: From the data given in the problem,

Initial velocity of the train u = 0 m/sec,

Acceleration a = 16.5 m/sec^2,

time t = 13.1 sec,

Velocity after 13.1 sec v=?

and the distance traveled S_1 =?

Substitute  the values in the equation v =u+at

we get v= 0+(16.5)(13.1) = 216.15 m/sec.

Substitute in equation S_1 = ut + \frac{1}{2} at^2

we get S_1 = (0)(13.1)+\frac{1}{2} (16.5)(13.1)^2

S_1 = 0+1415.78 =1415.78 m

After that the train travels with velocity v=216.15 m/sec for  69.7sec. calculate distance traveled S_2 during this time.

v=216.15 m/sec, t = 69.7 sec S_2=?

substitute the values in equation S_2= vt = (216.15)(69.7)=15065.66 m

Finally the trains decelerates  at the rate of 3.45 m/sec^2 and comes to rest.find distanceS_3 traveled before coming to rest.

Acceleration a = -3.45 m/sec^2 ,

Initial velocity u = 216.15m/sec,

Final velocity v=0,

distance traveled S_3 =?

substitute the values in v^2 - u^2 = 2as,

we get 0^2 - 216.15^2 = 2(-3.45) S_3

-6.90 S_3 = -46720.82

S_3 = 6771.13.

Total distance traveled by train S = S_1 +S_2+ S_3 = 1415.78+15065.66+6771.13=23252.57 m  or 23.252 km.

Kinematics

1.The terms Rest & Motion are relative terms.

  • There is no object in the universe  which is at absolute rest.Ex: A text book in a book rack may be  relatively at rest, with respect to the immediate surroundings. But, the earth is revolving  around sun, therefore every particle on earth will be in motion including the text book.Hence, the text book  also will be in motion with respect to Sun.
  • There is no object in the universe which is in absolute motion. Ex:A person traveling in a  train will be in motion as the train is moving.But, with respect to the fellow passengers and all the non moving objects in the train the person will be relatively  at rest.

2. The shortest distance between initial and final positions of a moving body is called its displacement.

  • Ex: If  a object starts its journey from a point A, travels in a circular path and again reaches the point A. Then its displacement is Zero,because its  initial and final points are same. But, the distance  traveled  is d = 2\pir = length of the circumference of circular path, i.e the distance (d) traveled by a object may not be equal to its displacement (S). When the object moves in a curved path or zig zag path d>S.
  • If the object travels in a specified direction along straight line path from a point A to a point B. In this case the distance d=AB and displacement S=AD will have same magnitudes i.e d=S.
  • Displacement is a vector and distance is a scalar quantity.
  • An object started from rest,travels with uniform acceleration.Find the ratio of its displacements in 1^{st},2^{nd},.  . .   .   .  n^{th} seconds of its journey. s_1:s_2:s_3: .  .  .  .  .  .  .  . :s_n= 1:3:5: .   .   .  .  .  .  .  .  . : (2n-1).
  • If body starts with an initial velocity u, and travels  with uniform acceleration  a .Find the ratio of its displacements in 1^{st},2^{nd},.  . .   .   .  n^{th} seconds of its journey. s_1:s_2:s_3: .  .  .  .  .  .  .  . :s_n= (2u+a) : (2u+3a) : (2u+5a) : .   .   .  .  .  .  .  .  . : {2u+(2n-1)a}.

Speed (v) : The distance traveled by a body in unit time is called Speed of the body.Speed is a scalar quantity.

Speed (v) = \frac{Distance}{Time}

Velocity (v) : Displacement of a body in unit time is called Velocity (or) Rate of change of displacement of a body is called its velocity. Velocity is a vector .

Velocity (v) = \frac{Displacement}{Time}

  • Units: Both speed and velocity have same units.  C.G.S  unit is cm / sec;, F.P.S unit is ft / sec; & M.K.S (or) S.I unit is  m / sec.
  • Relation between magnitude of velocity and speed will be speed \geq velocity.
  • For a object moving in a circular path with a constant speed, the magnitude of velocity remains constant.But,the direction of motion the object  continuously changes.
  • Velocity of an object is said to be constant if  a)its magnitude of velocity is constant  and b)its direction of motion remains unchanged.
  • If the velocity of an object is varying  (not constant) such object will possess acceleration, i.e the condition for an object to possess acceleration is  v\neq constant.
  • An object started from rest and traveling with uniform acceleration a.Find the ratio of velocities at the 1^{st},2^{nd},.  . .   .   .  n^{th} seconds of its journey.v_1:v_2:v_3: .  .  .  .  .  .  .  . :v_n= 1:2:3: .   .   .  .  .  .  .  .  . : n.
  • If body starts with an initial velocity u, and travels  with uniform acceleration  a .Find the ratio of its velocities at 1^{st},2^{nd},.  . .   .   .  n^{th} seconds of its journey. v_1:v_2:v_3: .  .  .  .  .  .  .  . :v_n= (u+a) : (u+2a) : (u+3a) : .   .   .  .  .  .  .  .  . : (u+na).

Acceleration ( a ) : Rate of change of velocity of a body is known as its acceleration. (or) Change in velocity of a body in unit time is called its  acceleration. Acceleration is a vector.

  • Acceleration in a body may be due to a) change in magnitude of velocity of a body (or) b) change in direction of velocity of the body or   c) change in both magnitude and direction of velocity of the body.
  • Ex: For a vertically projected body or for a freely falling body the acceleration is due to change in the magnitude of velocity.
  • Ex:For a body moving along a circular path with constant speed, the acceleration is due to change in the direction of velocity .
  • Units : C.G.S unit is cm/ sec^2 ; F.P.S unit is ft/sec^2 and M.K.S or S.I unit is m/sec^2 .
  • Direction of acceleration : a) For freely falling body the direction of  velocity and acceleration are same, in this case acceleration is considered to be positive( + )   b) For a body moving up vertically , direction of acceleration will be opposite to direction velocity of the body,in this case the acceleration is considered to be negative ( – )    c)For the body in circular motion the direction of acceleration will be perpendicular to the direction of velocity at every point.
  • Acceleration of a body may not be zero, even if the velocity of the body is zero. Ex: For a vertically projected body , when it is at Maximum height its velocity becomes zero but its acceleration is equal to ” g “.

Freely falling body : Any object falling under the influence  of gravitational force, with acceleration due to gravity is called freely falling body.

  • Ratio of displacements  of a freely falling body after 1sec,2sec, . . . . . . . . . . . . . . . . n sec of its journey will be  1 : 4 : 9 : 16  . . . . . . . . . n^2.
  • Ratio of displacements of a freely falling body 1st,2nd,3rd  …… nth sec of its journey will be 1 : 3 : 5 : 7 : 9 . . . . . . . . (2n-1).
  • Ratio of velocity of a freely falling body after 1 sec, 2 sec, 3sec , . . . . . . . . . . . . . .  of its journey will be 1 : 2 : 3 : 4 :
  • Ratio of velocity of a freely falling body after 1st , 2nd , 3rd , 4th . . . . . . . . . . . . . . . . .    of its journey will be 1 : 1 : 1 : . . . . . . . . . .   i.e its velocity in every interval is equal to “g”.
  • For a freely falling body velocity will continuously change (Increases) and the acceleration ( g ) remains constant.

Vertically projected up body: For a body projected vertically up with an initial velocity u,

  • Velocity at maximum height H_max will be zero.
  • At any point during its upward journey, direction of velocity will be opposite to direction of acceleration.
  • Velocity of the object at any point in its path is same in magnitude,when it is going up and coming down.
  • It reaches ground with the velocity(u) with which it is projected.
  • Displacement in nth second of its up ward journey S_n= u-\frac{n^2 - (n-1)^2}{2}g
  • For a vertically projected up body velocity will continuously change (decrease) and the acceleration ( -g ) remains constant.
  • Ratio of velocity of a freely vertically projected up body after 1 sec, 2 sec, 3sec , . . . . . . . . . . . . . .  of its journey will be  v_1:v_2:v_3 . . . . . . . . . . . .  :v_n  = (u-g) : (u-2g) : (u-3g) : . . . . . . . . . . . . . .  : (u-ng).

Projectile : A body projected with an angle other than  90^0 (≠ 90^0) is called a projectile .

  • Path of a projectile is parabola.
  • For a projectile as no force acts on it in the horizontal direction, it does not possess  acceleration in horizontal direction.
  • The horizontal component of velocity remains constant through-out its journey.
  • gravitational force acts on it vertically downwards,hence it will possess acceleration   equal to acceleration due to gravity and vertical component of velocity will be different at different points.
  • Path of a body projected horizontally from top of a tower  is a parabola.
  • Path of  an object dropped from aeroplane flying at certain height will be a) parabola with respect to a stationary  observer, but b) path of that object is a straight line with respect to an observer in the aeroplane.
  • If a body is projected horizontally from the top of a tower and another body dropped freely from the same height at the same time, both will reach the ground at the same time.
  • At maximum height of a projectile : a)It will possess only horizontal velocity b)Horizontal component of velocity v_x = u cos\theta c)vertical component of velocity is = 0 d) K.E = \frac{1}{2} mu^2cos^2\theta  e)P.E = \frac{1}{2} mu^2sin^2\theta  f) velocity is minimum, hence K.E also will be minimum    G) P.E is maximum as it is at maximum height.


Yahoo Answers.

Q: A particle is thrown vertically upwards with speed 20 meters per second. Two seconds later, another particle is thrown vertically upwards from exactly the same place. Its initial velocity is also 20 meters per second. How long after the first particle was projected, do the two particles collide? ( Question asked by gloom striken in Yahoo answers).

Ans : Initial velocity of the 1st particle u_1=20 m /sec assume acceleration due to gravity g = 10m/sec^2.

Maximum height traveled by this particle H_{max}= \frac{u_1^2}{2g} = \frac{20^2}{2\times10}.

H_{max}= \frac{400}{20} ; H_{max}= 20m.

Time of ascent t_a= u/g =20/10 =2 sec.

i.e when the second object is projected from ground, first object is at maximum height .

When a body is dropped from a  Maximum height or tower of height h and another body is thrown up vertically with a velocity u then they will meet after t=h/u seconds.

t =20/20=1 sec

They will collide after 1sec  after the II particle is projected or 3sec after the I particle is projected.

Important formulas in Physics.

Kinematics :

    • Average Speed = \frac{Totaldistancetraveled}{Total time taken}
    • If the body covers 1st half of distance with a speed x and the second half with a speed y,then the average speed = \frac{2xy}{x+y}
    • If the body covers 1st 1/3rd of a distance with a speed x , and 2nd 1/3 with a speed y , and the 3rd 1/3rd distance with a speed z, then average speed =\frac{3xyz}{xy+yz+zx}
    • Average velocity = \frac{TotalDisplacement}{Total time taken}
    • If a body travels a displacement s_1 in t_1 seconds and a displacement s_2 in t_2 seconds, in the same direction then   Average velocity = \frac{s_1+s_2}{t_1+t_2} .
    • If a body travels a displacement s_1 with velocity v_1 , and displacements_2 with velocity v_2 in the same direction then Average velocity = \frac{( s_1 + s_2) v_1v_2}{s_1v_2+s_2v_1}
    • If a body travels first half of the displacement with a velocity v_1and next half of the displacement with a velocity v_2 in the same direction , then                                                          Average velocity = \frac{2v_1v_2}{v_1+v_2} .
    • If a body travels a time t_1 with velocity v_1 and for a time t_2 with a velocity v_2 in the same direction then Average velocity = \frac{v_1t_2+V_2t_1}{t_1+t_2} .
    • If the body travels 1st half of the time with a velocity v_1 next half of the time with a velocity v_2 in same direction , then  Average velocity = \frac{v_1+v_2}{2}
    • For a body moving with uniform acceleration if the velocity changes from u to v in t seconds, then Average velocity = (u+v)/2 .
  • Equations of motion of a body moving with uniform acceleration along straight line.
  • a) V=u+at    b) S=ut+\frac{1}{2}at^2  c) v^2u^2 =2as
    • Distance traveled  in the nth second    s_n=u+a(n-1/2)
    • Equations of motion for a freely falling body ( Note: we can obtain these equations by substitution of u=0 and a=g in above equations . a) v=gt b) S=1/2 gt^2 c) v^2 = 2gs and the equation for the distance traveled in nth second changes to          s_n=g(n-1/2)
    • Equations of motion of a body projected up vertically :(we will obtain these equations by substitution a=-g in equations of motion)
    • a) v=u-gt  b) S=ut-1/2gt^2  c) v^2-u^2=-2gs and s_n=u-g(n-1/2)
    • Equation for maximum height reached H_{max} = \frac{u^2}{2g} \Rightarrow H_{max} \alpha u^2
    • Time of ascent t_a=\frac{u}{g};     t_a\alpha u
    • Time of descent  t_d =\frac{u}{g}t_d \alpha u
    • Time of flight T=2u/g
    • When a body is thrown up from top of a tower or released from a rising baloon,with velocity u.Displacement traveled before reaching ground              S=-ut+1/2gt^2. (t= time during which the object is in the air and S=h=height of the tower).
    • When a body is dropped from a tower of height h and another body is thrown up vertically with a velocity u then they will meet after t=h/u seconds.
    • When a body is dropped from a tower of height h . Its velocity when it reaches ground v=\sqrt(2gh)
    • If the displacements of a body in m^{th} ,n^{th} seconds of its journey.Then the uniform acceleration of the body a=\frac{s_n-s_m}{n-m}
    • From the above equation we can observe that by substituting n=1,2,3,4,…. we get a=s_2-s_1=s_3-s_2=……….. =s_{n-s}s_{n-1} =a.
    • A body projected up with velocity u from the top  of a tower reaches ground in t_1 seconds.If  it  is thrown down with the same velocity u it reaches ground in t_2 seconds.Then, when it is dropped freely the time taken to reach the ground will be t=\sqrt(t_1t_2)   and h=1/2 gt_1t_2   and t_1 -t_2 =2 \frac{u}{g}.
    • Projectile motion :Let us suppose that a projectile is projected with an initial velocity u making an angle \theta with x axis. a)Horizontal component of velocity u_x = u cos\theta , and u_x =v_x ,which will be constant through out the flight of the projectile as horizontal component of acceleration a_x = 0.  b)Vertical component of velocity of the projectile u_y = u sin\theta. Vertical component of velocity at any time of its journey v_y =u_y-gt or v_y =usin\theta -gt.   c)Magnitude of the resultant velocity V = \sqrt(v_x^2+v_y^2)  and the angle x made by v with the horizontal is given by Tan\alpha= \frac{v_y}{v_x}
    • Time of ascent = \frac{usin\theta}{g}
    • Time of descent = \frac{usin\theta}{g}
    • Time of flight = \frac{2u sin\theta}{g}
    • Maximum height reached  H_{max} =\frac{u^2 sin^2\theta}{2g} ;\frac{H1}{H2} =\frac{sin^2\theta_1}{sin^2\theta_2} when u is same
    • Horizontal Range R= \frac{(u^2sin2\theta)}{g }=\frac{2u^2 sin\theta cos\theta}{g}  a)R is maximum when \theta =45^0  b)R_{max} =\frac{u^2}{g}  c) If T is the time of flight, R=u cos\theta\times t  d)For given velocity of projection R is same for the angles of projections \theta and (90-\Theta)  Ex: 25 and 65 i.e the Range for those two angles will be same whose sum is 90^0)

Definitions -Kinematics

Kinematics : The study of motion of bodies, does not taking in to consideration the cause  for motion is called Kinematics.

Rest: If the position of an object remains to be same for any length of time with respect to its surroundings, then the object  will be at rest with respect to those surroundings.

Motion: If the position of  a body is continuously changing with respect its surroundings with time , then the body is said to be in motion.

Uniform motion : If a body moving  along a straight line path travels equal distances in equal interval of time, then the body is said to be in uniform motion.

Non Uniform motion : If a body moving along straight line travels unequal distances in equal intervals of time, the body is said to be in Non uniform motion. (or) If  a body has unequal displacements  in in equal  intervals of time, it will be in Non uniform motion.

Displacement ( S ) : The change in the position of a body in a specific direction, is called Displacement. Displacement is a vector quantity.

Speed (v) : The distance traveled by a body in unit time is called Speed of the body.Speed is a scalar quantity.

Speed (v) = \frac{Distance}{Time}.

Uniform speed : If the distance traveled by a body  in equal intervals of time are same, for any length of time then the body is said to be in uniform speed.

Non-uniform speed : If the distance traveled by a  body in equal intervals of time are different,the body is said to be in non uniform speed.

Velocity (v) : Displacement of a body in unit time is called Velocity (or) Rate of change of displacement of a body is called its velocity. Velocity is a vector .

Velocity (v) = \frac{Displacement}{Time}.

Uniform velocity (v) : If the displacements  of a body  in equal intervals of time are same, for any length of time then the body is said to be in uniform motion.

Non-uniform velocity : If the displacements of a body in equal intervals of time are different,the body is said to be in non uniform motion.

Instantaneous Velocity: The velocity of a body at a particular instant of time is known as  instantaneous  velocity.

Average Velocity : If the displacement of a body in a small interval of time dt  is  ds then, the ratio of the displacement ds to the time interval dt is called average velocity. Average velocity v = (ds)/dt.

Initial velocity ( u  ) : Velocity of the body in the beginning of the observation is called initial velocity.

Final velocity ( v ) : Velocity of the body at the end of the observation is called its final velocity.

Acceleration ( a ) : Rate of change of velocity of a body is known as its acceleration. (or) Change in velocity  of a body in unit time is called its  acceleration. Acceleration is a vector.

Deceleration : If the velocity of a body is continuously decreasing with time, the body will possess negative acceleration. It is also called deceleration.

Uniform acceleration: If the change (increase  or decrease ) in velocity of a body is constant in equal intervals of time, the body will possess uniform acceleration.

Non uniform acceleration : If the change (increase  or decrease ) in velocity of a body is different  in equal intervals of time, the body will possess non-uniform acceleration.

Acceleration due to gravity ( g ): The acceleration in a freely falling body due to gravitational  force  acting  on it is called acceleration  due  to  gravity. This is a vector. The value of  g  in C.G.S system  is   980cm sec^{-2}  and in M.k.s & S.I systems is  9.8m sec^{-2}  .

Freely falling body : Any object falling under the influence  of gravitational force, with acceleration due to gravity is called freely falling body.

Maximum Height ( H) : The distance traveled by a  body vertically projected up, just before coming to rest is called, maximum height.

Time of ascent (t_a  ) : The time taken by  a vertically projected body to reach the maximum height is called time of ascent.

Time of descent ( t_d ) : The time taken by a body to reach the ground from its maximum height is called time of descent.

Time of flight ( T ) :The total time during which the vertically projected body will remain in air is called time of flight.

( or ) Total time taken by a projectile to reach the same horizontal plane from which it is projected is called called time of flight.

Time of flight  T =( t_a ) +  ( t_d ) .

Projectile : A body projected with an angle other than  90^0 (≠ 90^0) is called a projectile .

Horizontal Range ( R ) :Maximum horizontal distance traveled by the body before it touches the ground (or) Maximum horizontal distance traveled by the body before it touches the point on the same level of projection is called horizontal range ( R ).


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