# 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 +g$t_a$

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.
.

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 =g$t_d$,

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.
.

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=9.8$m/sec^2$; 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=9.8$m/sec^2$,

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,

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 3$msec^{-2}$ 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= 3$msec^{-2}$,

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 4$msec^{-2}$.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=4$msec^{-2}$,

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=30$msec^{-2}$.

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 2.5$msec^{-2}$ 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 = 2.5$msec^{-2}$,

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= 40$m/sec^2$

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_4$$s_2$ = a(4-2) =2a

therefore 2a = $s_4$$s_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 $S_2$ = u$t_2$+$\frac{1}{2} at_1^2$,

$S_2$ = 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+a$t_2$

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 16.5$m/sec^2$  for  13.1 sec.It runs at constant speed for 69.7s and slows down at a rate of 3.45 $m/sec^2$until 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 distance$S_3$ 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.

# Question By Creev in Yahoo answers

Q:A car accelerates uniformly from rest to 20.0 m/s in 5.6 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 9.0 x 10^3 N, and (b) the weight of the car is 1.4 x 10^4 N.

Ans : Read the problem note the given physical quantities with symbols from it. Given that Initial velocity u=o m/sec , Final velocity v=20m/sec , Time  t = 5.6 sec , let g=10m/$sec^2$ .

Acceleration of the car a =(v-u)/t =(20-0)/5.6 =(20/5.6)m/$sec^2$

a) Mass of the car  m= $\frac{Weight}{g}$

m=(9.0 x $10^3$ )/10 ; m = 900 kg

Force on car F = $m\times a$ =$900\times\frac{20}{5.6}$ = $\frac{18000}{5.6}$ N = 3214.28N

Average power P = $F\times V$ ; P= $3214.28\times 20$ = 64285.6watt  or 64.2856 kwatt.

b)Mass of the car m = (1.4 x $10^4$ )/10 =1400kg

Force on car F = $m\times a$ =$1400\times\frac{20}{5.6}$ = $\frac{28000}{5.6}$ N = 5000N

Average power P = $F\times V$ ; P= $5000\times 20$ = $10^5$watt  or 100kwatt.

Q : Please define and/or explain impulse. Please say more than just product of F and t, and also the change of momentum. (Question by Tim C in Yahoo answers)

Ans : i) The effect of a large force acting on a body for a very short duration of time is called Impulse. ii) Change in momentum of a body is called Impulse iii ) The product of force applied on a body and the time for which it is applied is called Impulse. i.e Impulse = F x t (X is not vector product it is normal multiplication )

But if you try to understand the term Impulse from the equation Impulse = F x t  , you will be confused. Because from this equation we come to a conclusion that, Impulse is directly  proportional to time ( t ) i.e longer the period of time more will be the Impulse.Actually the Impulse will be less if the time period increases.

Smaller the time larger will be the impulse.

Ex: i)When a cricketer hits the ball hard, the time of contact (time of force applied ) of the ball is very, very short .Hence the effect will be large.More harder he hits the ball , the time of contact decreases further ,therefore the Impulse will further increase.

ii)when a person jumps on a hard surface from certain height,the moment his feet touches the floor they will be brought to rest  , i.e the time of application of force is very small.Hence, the effect(Impulse) is high.But, if he jumps from the same height in to sand the foot will come to rest after some time, in this case even though the force is same but the time of application of force is increased.Hence, the effect of the force(Impulse) will be less.