Question

If you throw a 2.5 kg ball straight down at 40 m/s from the top of a very high cliff, how fast will it be going when it’s 39 m below release point? Round your answer to the nearest tenth and include the appropriate unit.

Answer 1

Given

Mass of the ball,, m=2.5 kg

Initial velocity, u=40 m/s

Distance travelled, s=39 m

To find

Velocity when it has crossed 39 m.

Explanation

Let the required velocity be v

Thus, by equation of kinematics

$\begin{gathered} v^2-u^2+2gs \\ \Rightarrow v^2=40^2+2\times9.8\times39 \\ \Rightarrow v=48.62\text{ m/s} \end{gathered}$

Conclusion

The required velocity is 48.62 m/s