A body is projected upwards with a velocity of $98 m / s$ . The second body is projected upwards with the same initial velocity but after $4 s$ . Both the bodies will meet after

6 s

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8 s

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10 s

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12 s

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Solution

The correct option is D $12 s$
Let $t$ be the time of flight of the first body after meeting then $(t - 4) s$ will be the time of flight of the second body.
Since, $h_{1} = h_{2}$
$98 t - \frac{1}{2} g t^{2} = 98 (t - 4) - \frac{1}{2} g (t - 4)^{2}$
On solving we get,
$t = 12 s$

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A body is projected upwards with a velocity of 98 m/s. The second body is projected upwards with the same initial velocity but after 4 s. Both the bodies will meet after

The correct option is D 12 sLet t be the time of flight of the first body after meeting then (t−4) s will be the time of flight of the second body. Since, h1=h2 98t−12gt2=98(t−4)−12g(t−4)2 On solving we get, t=12 s

A body is projected upwards with a velocity of $98 m / s$ . The second body is projected upwards with the same initial velocity but after $4 s$ . Both the bodies will meet after

The correct option is D $12 s$
Let $t$ be the time of flight of the first body after meeting then $(t - 4) s$ will be the time of flight of the second body.
Since, $h_{1} = h_{2}$
$98 t - \frac{1}{2} g t^{2} = 98 (t - 4) - \frac{1}{2} g (t - 4)^{2}$
On solving we get,
$t = 12 s$