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A body P is thrown vertically up with velocity 30 ms^–1 and another body Q is thrown up along the same vertical line with the same velocity but 1 second later from the ground. When they meet (g = 10 ms^–2)

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P travels for 2.5s

Q travels for 3.5s

P travels for 3.5s

Q travels for 1s

answer is C.

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Detailed Solution

Assuming they meet at time t,

$\begin{array}{l} s_{1} = ut - \frac{1}{2} {gt}^{2} (for P); g = 10 {ms}^{- 2} \\ s_{2} = u (t - 1) - \frac{1}{2} g (t - 1)^{2} (for Q); s_{1} = s_{2} \\ \Rightarrow 30 t - \frac{1}{2} {gt}^{2} = 30 (t - 1) - \frac{1}{2} g (t - 1)^{2} \end{array}$

$\Rightarrow t = 3.5 s$
So P travels for 3.5 s

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