Question

A stone projected vertically up from the top
of a cliff reaches the foot of the cliff in 8s.
If it is projected vertically downwards with
the same speed, it reaches the foot of the
cliff in 2s. Then its time of free fall from
the cliff is​

Answer 1

EXPLANATION.

• GIVEN

stone project vertically up from the top of a cliff

reaches the foot of the cliffs = 8 seconds

From newton second law of kinematics.

$s = ut + \frac{1}{2}g {t}^{2}$

s =

$u(8) + \frac{1}{2} (g)(8) {}^{2}$

s = 8u - 32g ....(1)

it is projected vertically downwards with

the same speed, it reaches the foot of the

cliff in 2s

$s = ut + \frac{1}{2}gt {}^{2}$

$s = u(2) + \frac{1}{2}g(2) {}^{2}$

s = 2u - 2g .... (2)

From equation (1) and (2)

we get,

$t = \sqrt{ \frac{2h}{g} }$

6u = 30g

u = 5g

put u = 5g in equation (2)

2(5g) - 2g

10g - 2g = 8g = H

$t \: = \sqrt{ \frac{2 \: \times \: 8g}{g} } = \: 4 \: \: seconds$

TIME FOR FREE FALL FROM THE CLIFF IS = 4 SECONDS