Question

Ball A is droped from the top of a building. At the same instant ball B is thrown vertically upwards from
the ground. When the balls collide, they are moving in opposite directions and the speed of Ais twice
the speed of B. At what fraction of the height of the building did the collision occur?
A) 1/3
B) 2/3
C) 1/4
D) 2/5​

Answer 1

Explanation:

hey there !

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.

Hope this is helpful for you ❤️