Question

A diver running 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. how high was the cliff and how far from its base did the diver hit the water?

Answer 1

Answer:

a) Height of cliff = 44.145 meter

b) 7.5 meter far from its base the diver hit the water.

Explanation:

 We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Diver's vertical motion:

   Initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$, we need to calculate displacement when time is 3 seconds.

  $s= ut+\frac{1}{2} at^2\\ \\ =0*3+\frac{1}{2} *9.81*3^2\\ \\ =44.145meter$

 So height of cliff = 44.145 meter.

b) Diver's horizontal motion:

Initial velocity = 2.5 m/s, acceleration = 0 $m/s^2$, we need to calculate displacement when time is 3 seconds.

$s= 2.5*3+\frac{1}{2}*0*3^2\\ \\ =7.5meter$

So 7.5 meter far from its base the diver hit the water.