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 bywhich the person can enter the water. Suppose he enters the waterat a speed of 13 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 5.20 m above the water? Ignore the effects of air resistance.

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2abehn · Accepted Answer

In the first path, the initial speed of the person is 0 m/sand when he enters water his speed is 13 m/s.Initially he is at a height of h m w.r.t. water. His potentialenergy, U = mghWhen he reaches water his energy is gfully kinetic. So K.E., T=mv22By conservation of energy, △U=△Tmgh=mv22⇒h=v22g=1322×9.8=8.62 mHe jumps from a height of 5.2 m above water level. So, from the topof the clif the person will be at a depth of (8.62-5.20=3.42 m)when he releases the rope. Applying law of coservation of energyagain for the path 2,mgh′=mv′22⇒v′=(2gh′)12=(2×9.8×3.42)12=8.2 ms

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 path

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