A girl standing on a pedestal at rest throws a ball upwards with maximum possible speed of $50 m s^{- 1}$ . If the platform starts moving up at $5 m s^{- 1}$ and the girl again throws the similar ball in similar way. The time taken in the previous case and the time taken in the second case to return to her hands are

10 s, 5 s

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5 s, 10 s

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10 s, 10 s

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5 s, 5 s

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Solution

The correct option is C $10 s, 10 s$
Time taken to reach highest point $= t = \frac{u}{g} = \frac{50}{10} = 5 s$
Total time taken to reach back $= 5 + 5 = 10 s$
When the platform start moving with uniform speed, the acceleration is nil. Motion of the ball and the girl w.r.t each other is unaffected so total time taken in this case also is $10 s$

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