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Grade: 12
        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?
2 years ago

Answers : (1)

Arun
23328 Points
							
Dear Pavan
 
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
 
Regards
Arun (askIITians forum expert)
2 years ago
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