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?

Short answer

The height of the cliff from which the stone is dropped is 61.75 m.

Step-by-step solution

Step1. What is the acceleration due to gravity?

Acceleration due to gravity is the acceleration of an object falling under the influence of gravity (and no other force, including wind resistance). Its value is generally taken as$g=9.8{\mathrm{ms}}^{-2}$.

Step2. Given data and assumption

The time elapsed before the stone hit the ground,$t=10.4\mathrm{s}$

The initial velocity of the stone,$u=0$

Assume that the height of the cliff is h and acceleration due to gravity is $\mathrm{g}=9.8{\mathrm{ms}}^{-2}$.

Step3. Using the second equation of motion

From the second equation of motion, you can write the displacement of the stone as.

$h=u\times t+\frac{1}{2}\mathrm{g}{t}^2$

Substituting the values in the above equation, you get.

$h=0+\frac{1}{2}\times 9.8\times 3.{55}^2$

Solving the above equation for h, you get.

$h=61.75\mathrm{m}$

Thus, the height of the cliff is 61.75 m.