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Ad libitum [116K]
3 years ago
13

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, a

nd 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.
Physics
1 answer:
Degger [83]3 years ago
5 0

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

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