Question

The drawing 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 10.2 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 2.33 m above the water? Ignore the effects of air resistance.

Answer 1

1) path 1

applyingh energy conservation

initial velocity u = 0

so 1/2 m v² = mgh

1/2 m x10.2² = m g h

h = 5.3 m

path 2

let height of fall of person when he releasesthe rope is h' = 2.33 m

let velocity of person at the point of releaseis v'

then total energy is 1/2 m v'² +mgh'

conservation of energy

(1/2)mv'² + mgh' = mgh

1/2 m v'² + m g x2.33 =mg x 5,3

v' =7.63 m/s

answer