Question

The drawing below shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the water below. There are two paths by which the person can enter the water. Suppose he enters the water at a speed of 16.5 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 3.55 m above the water? Ignore the effects of air resistance.

Answer 1

Answer:

$v = 14.23 m/s$

Explanation:

As per energy conservation we can say that during path 1. his initial total mechanical energy must be equal to final total mechanical energy

So we will have

$mgH = \frac{1}{2}mv^2$

so we will have

$m(9.81)(H) = \frac{1}{2}m(16.5)^2$

$H = 13.88 m$

Now while he is moving on path 2. then again we can use energy conservation to find the speed

$mg(H - y) = \frac{1}{2}mv^2$

$m(9.81)(13.88 - 3.55) = \frac{1}{2}m v^2$

$v = 14.23 m/s$