Question

A ball 'A' is dropped from the top of a building and at the same time an identical ball 'B' is thrown vertically upward from the ground. When the balls collide, the speed of A is twice that of B. At what fraction of the height of the building did the collision occur?

Answer 1

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,
$v_A=2v_B$

√{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.

Answer 2

Explanation:

Let t = t_o be the time they need to meet. Then the dropped ball has the speed

vA = g * t_o

while the rising ball has speed

vB = u - g*t_o

We are given vB = 1/2 * v_A

substituting g * t_o for vA

we have vB = 1/2 * g*t_0

So far we have

vB = 1/2 * g*t_o

vB = u - g*t_o

By equality

1/2 * g*t_0 = u - g*t_o

solving for u

u = 3/2*g*t_0

Ball A fell

f = 1/2 g t_o^2

Ball B rose

r = u*t_0 - 1/2*gt_o^2

substitute u = 3/2*g*t_o

Ball B rose

r = (3/2*g*t_o) * t_o - 1/2*gt_o^2

r = g*t_o^2

the heigth (sic) is

h = r+f = g*t_o^2 + 1/2*g*t_o^2 = 3/2 gt_o^2 = 3/2 * r

So

h = 3/2 * r ---> r = 2/3 * h