Question

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is the height of the cliff?

Answer 1

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = $\frac{1}{2}mv^2$.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 $\frac{1}{2}mv^2=\ mgh$

                ⇒ $v^2=\ 2gh$

                ⇒ $h=\ \frac{v^2}{2g}$

                ⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

                ⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m