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valina
3 months 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:
Keith_Richards [3.2K]3 months ago
3 0

Answer:

11.56066 m/s

Explanation:

m = Mass of individual

v = Velocity of individual = 13.4 m/s

g = Gravitational acceleration = 9.81 m/s²

v' = Velocity of the individual after dropping

At the surface, kinetic and potential energy will equalize

\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

The cliff's height is 9.15188 m

Define fall height as 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 person's speed is 11.56066 m/s

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