First slide

Vertical projection from ground

Question

A body X is projected upwards with a velocity of $98 {ms}^{- 1}$ , after 4 s, a second body Y is also projected upwards with the same initial velocity. Two bodies will meet after

Moderate

A

8 s
B

10 s
C

12 s
D

14 s

Solution

Let t second be the time of flight of the first body after meeting, then (t $-$ 4) second will be the time of flight of the second body.
As the initial velocity at which the bodies A and B projected are same and also the position of meeting will be also same.
So, $h_{x} = h_{y}$
$∴ 98 t - \frac{1}{2} g t^{2} = 98 (t - 4) - \frac{1}{2} g (t - 4)^{2}$
On solving, t = 12 s

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