Question

A cliff diver dove off a cliff at 8 m/s and landed in the water 24 m away from the cliff. How high is the cliff?

Answer 1

We are given the following information

Horizontal distance covered = 24 m

Initial speed = 8 m/s

We are asked to find the height of the cliff.

Let us first find the time.

$\begin{gathered} v_{ix}=\frac{\Delta x}{t} \\ t=\frac{\Delta x}{v_{ix}} \end{gathered}$

Where △x is the horizontal distance covered and vix is the initial horizontal speed of the diver.

$t=\frac{24}{8}=3\; s$

So, the time is 3 seconds.

The vertical distance covered by the diver is given by

$y=\frac{1}{2}\cdot g\cdot t^2$

Where g is the acceleration due to gravity that is 9.81 m/s^2

$y=\frac{1}{2}\cdot9.81\cdot3^2=44.145\; m$

Therefore, the height of the cliff is 44.145 m