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.

Answer 2

Cliff is 61.81 m high.

Given :

Time, t = 3.55 sec

Solution :

We know that,

$\rm s = ut + \dfrac{1}{2}at^2$    ----- (1)

where,

s is the displacement,

initial velocity - u = 0,

t is the time and

a is the gravitational acceleration which is $\rm 9.81 \; m/sec^2$.

Now put the values of a, t and u in equation (1) we get,

$\rm s = 0 + \dfrac{1}{2}\times 9.81 \times (3.55)^2$

$\rm s = 61.81\;m$

Cliff is 61.81 m high.

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