Question

) a stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.55 s. How high is the cliff?

Answer 1

We use kinematic equation of motion,

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

Here, s is distance traveled, u is the initial velocity, t is time taken and g is acceleration due to gravity.

Given, $t =3.55 s$.

We take,  $u = 0$ because stone has no initial velocity.

Therefore,

$S = 0 \times t + \frac{1}{2} \times 9.8 \ m/s^2 \times (3.55)^2 s = 61.75 \ m$

Thus,  the height of cliff is approximately 61.75 m.