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