A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff upward with a velocity of $100 m s^{- 1}$ . At what time will the bullet and the block meet ? Also, where was the block with respect to the top of cliff when they met?

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

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

Open in App

Solution

The correct option is D
1s

Height of the tower $= s_{1} + s_{2} = 100 m$ .
where $s_{1}$ and $s_{2}$ are the distances covered by wooden block and bullet respectively.

Before the collision:
1. Distance covered by the block:
$s_{1} = \frac{1}{2} g t^{2} (∴ u = 0, a = g) = \frac{1}{2} \times 9.8 \times t^{2} = 4.9 t^{2}$

2. Distance covered by the bullet:
$s_{2} = u t - \frac{1}{2} g t^{2} = 100 t - \frac{1}{2} \times 9.8 \times t^{2} = 100 t - 4.9 t^{2}$

Since, $s_{1} + s_{2} = 100 m$
$4.9 t^{2} + 100 t - 4.9 t^{2} = 100 \Rightarrow 100 t = 100 \Rightarrow t = 1 s .$

And, distance covered by the block = $4.9 t^{2} = 4.9 \times (1)^{2}$
= 4.9 m.

Suggest Corrections

Similar questions

A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff upward with a velocity of $100 m s^{- 1}$ . At what time will the bullet and the block meet ? Take g= $9.8 m s^{- 2}$ .

Q. A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff in the upward direction with a velocity of

100 m s^{- 1}

. At what time will the bullet and the block meet ?

A wooden block is dropped from the top of a cliff 100 metre high and simultaneously a bullet of mass 10 gram is fired from the foot of the cliff upwards with a velocity of 100 m/s . The bullet and wooden block will meet each other after a time

Q. A wooden block of mass