Question

A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3s later. How high was the cliff and how far from its base did the diver hit the water?work needed

Answer 1

We use kinematic equation,

$h=ut+\frac{1}{2} gt^2$

Here, h is vertical height, u is initial vertical velocity, t is time taken and g is acceleration due to gravity.

As diver dives out horizontally, his velocity is directed horizontally; that is, the initial vertical velocity is 0. So above equation becomes

$h=\frac{1}{2} gt^2$

Given, t =3 s.

Therefore,

$h=\frac{1}{2}\times 9.8m/s^2(3\ s)^2 =0.5\times 9.8 m/s^2\times 9 s^2\\\\h=44.1\ m$.

Now the horizontal distance of the diver to hit the water from base,  

$=1.8\ m/s \times 3\ s=5.4\ m$