Question

A 271-gram ball is thrown at a speed of 36.2 m/s from the top of a 43.5-m high cliff. Determine the impact speed of the ball when it strikes the ground. Assume negligible air resistance.Answer: _________ m/s (round to the nearest hundredth)

Answer 1

The mechanical energy of the ball is conserved because the air resistance is negligible.

When the ball is thrown with an initial speed v₀, it is located at a height h. Then, its total mechanical energy, which is the sum of its kinetic and potential energies, is:

`E=(1)/(2)mv_0^2+mgh`

When the ball strikes the ground, all of its potential energy is converted to kinetic energy. Then, its mechanical energy at that point depends on the speed of the ball only:

`E=(1)/(2)mv^2`

Then:

`(1)/(2)mv^2=(1)/(2)mv_0^2+mgh`

Isolate v from the equation:

`\begin{gathered} \Rightarrow v^2=v_0^2+2gh \\ \\ \Rightarrow v=√(v_0^2+2gh) \end{gathered}`

Replace v₀=36.2m, g=9.81m/s^2 and h=43.5m:

`\begin{gathered} v=\sqrt{(36.2(m)/(s))^2+2(9.81(m)/(s^2))(43.5m)} \\ \\ =46.5178...(m)/(s) \\ \\ \approx46.52(m)/(s) \end{gathered}`

Therefore, the impact speed of the ball is approximately 46.52 m/s.