/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 The \(0.5-\mathrm{kg}\) ball is ... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 75

The \(0.5-\mathrm{kg}\) ball is fired from the tube at \(A\) with a velocity of \(v=6 \mathrm{~m} / \mathrm{s}\). If the coefficient of restitution between the ball and the surface is \(e=0.8\), determine the height \(h\) after it bounces off the surface.

Short Answer

Expert verified
The height \(h\) after the ball bounces off the surface is calculated to be approximately \(1.176 \mathrm{~m}\).

Step by step solution

01

Calculate the velocity after impact

The velocity after the impact can be calculated by multiplying the initial velocity with the coefficient of restitution. So, \(v'= e * v = 0.8 * 6 = 4.8 \mathrm{~m/s}\)
02

Determine the maximum height

Now to find the height, use the principle of conservation of energy. Initially, the total energy of the ball is all kinetic energy (since it hasn't risen yet), which equals \(0.5*m*(v')^2\), and at the maximum height after the bounce, its energy is all potential energy (since it has stopped moving), which equals \(m*g*h\). Setting these two equal gives \((1/2) * 0.5 * (4.8)^2 = 0.5 * 9.81 * h\). Solving for \(h\) will give us the height reached by the ball in meters after the bounce.

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Most popular questions from this chapter

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