Question

A 221-gram ball is thrown at a speed of 36.7 m/s from the top of a 39.8-m high cliff. Determine the impact speed of the ball when it strikes the ground. Assume negligible air resistance.

Answer 1

Given:

The mass of the ball is

$\begin{gathered} m=221\text{ g} \\ =0.221\text{ kg} \end{gathered}$

The initial height of the ball is

$h=39.8\text{ m}$

The initial speed of the ball is

$v_i=36.7\text{ m/s}$

To find:

the impact speed of the ball when it strikes the ground

Explanation:

The initial potential energy of the ball is

$\begin{gathered} (PE)_i=mgh \\ =0.221\times9.8\times39.8 \\ =86.2\text{ J} \end{gathered}$

The initial kinetic energy is

$\begin{gathered} (KE)_i=\frac{1}{2}mv_i^2 \\ =\frac{1}{2}\times0.221\times(36.7)^2 \\ =148.8\text{ J} \end{gathered}$

The final energy of the ball is fully kinetic energy. Let the final impact speed of the ball is

$v_f$

We can write, using the energy conservation principle that

$\begin{gathered} (PE)_i+(KE)_i=\frac{1}{2}mv_f^2 \\ 86.2+148.8=\frac{1}{2}\times0.221\times v_f^2 \\ v_f^2=2\times\frac{86.2+148.8}{0.221} \\ v_f=46.1\text{ m/s} \end{gathered}$

Hence, the final impact speed of the ball is 46.1 m/s.