Question

From the top of a building of height H, a Ball A is dropped from rest. Exactly at the same instant another ball B is thrown up vertically from the ground such that it travels along the same path as A. Both balls A and B collide in air after some time. At the time of collision it is observed that ball A has twice the speed of B. If the collision occurs at height h, find the value of h/H?


PLZ ANSWER ASAP

Answer 1

Answer:

Ball

is dropped from the top of a building. At the same instant ball

is thrown vertically upwards from the ground. When the balls collide, they are moving in opposite direction and the speed of

is twice the speed of

. At what fraction of the height of the building did the collision occurs ?

Answer 2

Answer:

Let height of building is h and x is the desired fraction.

for ball A,

initial velocity of ball A , u = 0

distance covered by ball A = (1 - x)h

so, use formula, S = ut + 1/2at²

(1 - x)h = 0 + 1/2 × g × t²

(1 - x)h = gt²/2

t = √{2(1 - x)h/g}

velocity at t sec , v = u + at

v = 0 + g√{2(1 - x)h/g} = √{2g(1 - x)h}......(1)

for ball B,

distance covered by ball B , S = xh

use formula, S = ut + 1/2at²

xh = u√{2(1-x)h/g} - 1/2 × g × (√{2(1-x)h/g})²

after solving , we get, u = √{gh/2(1 - x)}

velocity at t sec., v = u + at

= √{gh/2(1 - x)} -g√{2(1 -x)h/g}

= √{gh/2(1 - x)} - √{2g(1 - x)h} .......(2)

A/C to question,

√{2g(1-x)h} =2[√{gh/2(1 - x)} - √{2g(1 - x)h} ]

3√{2g(1 - x)h} = 2√{gh/2(1 - x)}

9(1 - x) = 1/(1 - x)

9(1 - x)² = 1

x = 2/3.

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