Question

A stone is dropped from the top of a cliff 40 m high and at the same instant another stone is shot vertically up from the foot of the cliff with a velocity 20 m per sec. Both stones meet each other after

पर्याय

• `1/2` sec

• `3/2` sec

• 2 sec

• 4 sec

Answer 1

2 sec

Explanation:

Height - covered by the projectile from the ground

`h = ut - 1/2 g  t^2`

`h = 20  t - 1/2 xx 9.8  t^2`

`h = 20  t - 4.9  t^2`  ......(i)

Distance covered by stone when dropped from the 40 m height at same time.

∴ `40 - h = 4t + 1/2 g  t^2`

⇒ `40 - h = 0 + 1/2 9.8  g  t^2`

From (i)

⇒ `40 - (20t - 4.9t^2) = 4.9  t^2`

⇒ `40 - 20t = 0`

⇒ t = 2 sec