# Consider a ball with mass m = 50 kg. The ball is at rest on a cliff 328 feet above the ground. ...

## Question:

Consider a ball with mass m = 50 kg. The ball is at rest on a cliff 328 feet above the ground.

(A) Derive an equation that describes the potential energy of the ball as a function of time after it falls off the cliff.

(B) How much kinetic energy (in Joules) does the ball have 2.1 seconds after it falls?

## Potential and Kinetic Energy:

The potential energy is the energy due to the position of a particle while the kinetic energy is the energy possessed by a particle due to the motion.

{eq}\begin{align*} PE &= mgh\\ KE &= \frac{1}{2}m{v^2} \end{align*} {/eq}

Given Data

• The mass of ball is: {eq}m = 50\;{\rm{kg}} {/eq}
• The ball is at rest on a cliff at {eq}328\;{\rm{ft}} {/eq} above the ground.

(A)

Calculate the vertical position of the ball from the ground.

{eq}h = {y_0} - \dfrac{1}{2}g{t^2} {/eq}

Here, the initial vertical position of the ball is {eq}{y_0} {/eq}, acceleration due to gravity is {eq}g {/eq} and the time is {eq}T {/eq}.

Calculate the potential energy of the ball.

{eq}\begin{align*} PE &= mgh\\ PE &= mg\left( {{y_0} - \dfrac{1}{2}g{t^2}} \right) \end{align*} {/eq}

Thus, the equation that describes the potential energy of the ball as a function of the time is

{eq}PE = mg\left( {{y_0} - \dfrac{1}{2}g{t^2}} \right). {/eq}

(B)

Calculate the speed of the ball after time t.

{eq}\begin{align*} v &= {v_0} + gt\\ v &= 0 + \left( {9.81\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right)\left( {2.1\;{\rm{s}}} \right)\\ v &= 20.601\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

Calculate kinetic energy of ball after 2.1 second of fall.

{eq}\begin{align*} KE &= \dfrac{1}{2}m{v^2}\\ KE &= \dfrac{1}{2}\left( {50\;{\rm{kg}}} \right){\left( {20.601\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)^2}\\ KE &= 10601.03\;{\rm{J}} \end{align*} {/eq}

Thus, the kinetic energy of ball after 2.1 second of fall is {eq}10601.03\;{\rm{J}} {/eq}.