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Q.

A ball is projected vertically upward with an initial velocity of 50 ms^–1 at t = 0s. At t = 2s. Another ball is projected vertically upward with same velocity. At t =________s, second ball will meet the first ball (g =10 ms^–2).

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answer is 6.

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Detailed Solution

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Let they meet at t = t
So first ball gets t sec. & 2^nd gets (t – 2) sec. & they will meet at same height
$\begin{array}{l} h_{1} = 50 t - \frac{1}{2} {gt}^{2} \\ h_{2} = 50 (t - 2) - \frac{1}{2} g (t - 2)^{2} \\ h_{1} = h_{2} \\ 50 t - \frac{1}{2} {gt}^{2} = 50 (t - 2) - \frac{1}{2} g (t - 2)^{2} \\ 100 = \frac{1}{2} g [t^{2} - (t - 2)^{2}] \\ 100 = \frac{10}{2} [4 t - 4] \end{array}$
5 = t – 1
t = 6 sec.

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A ball is projected vertically upward with an initial velocity of 50 ms–1 at t = 0s. At t = 2s. Another ball is projected vertically upward with same velocity. At t =________s, second ball will meet the first ball (g =10 ms–2).