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`        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

Arun
23751 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

```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions