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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12 s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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,
$t = 12 s$

Suggest Corrections

Similar questions

Q. 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

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

Q. 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

Q. A body A is projected upwards with a velocity of

98 \frac{m}{s}

. The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

Q. A body

X

is projected upwards with a velocity of

98 m s^{- 1}

, after

4 s

, a second body

Y

is also projected upwards with the same

Y

is also projected upwards with the same initial velocity. Two bodies will meet after

Join BYJU'S Learning Program

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,
$t = 12 s$