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Free_Kalibri [48]
3 years ago
7

Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. 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 13.4 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 2.34 m above the water? Ignore the effects of air resistance.
Physics
1 answer:
Alex73 [517]3 years ago
3 0

Answer:

11.56066 m/s

Explanation:

m = Mass of person

v = Velocity of person = 13.4 m/s

g = Acceleration due to gravity = 9.81 m/s²

v' = Velocity of the person in the second

The kinetic and potential energy will balance each other at the surface

\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m

Height of the cliff is 9.15188 m

Let height of the fall be h' = 2.34 m

\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s

The speed of the person is 11.56066 m/s

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