The old rubber boot shown in Figure 12.27 has two leaks. To what maximum height can the | StudySoup

Textbook Solutions for College Physics for AP® Courses

Chapter 12 Problem 20

Question

The old rubber boot shown in Figure 12.27 has two leaks. To what maximum height can the water squirt from Leak 1? How does the velocity of water emerging from Leak 2 differ from that of leak 1? Explain your responses in terms of energy.

Solution

Step 1 of 2

First we need to calculate the maximum height to which the water can squirt from leak 1.

Water from leak 1 exit to follow the path like a projectile motion from below mentioned formula of projectile for attain a maximum height.

Calculate maximum height of water squirt from leak 1.

\(h=\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Here, h is the maximum height, u is velocity of water at exit of squirt, and \(\theta\) is the angle.

For maximum height water exit at angle of.

Substitute \(45^{\circ}\) for \(\theta\)

\(\begin{array}{l}
h=\frac{u^{2} \sin ^{2} 45^{\circ}}{2 g} \\
=\frac{u^{2}}{2 g}
\end{array}\)

Hence, the maximum height to which water can squirt from leak 1 is

\(\frac{u^{2}}{2 g}\)

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full solution

Title College Physics for AP® Courses 1 
Author Gregg Wolfe, Irina Lyublinskaya, Douglas Ingram
ISBN 9781938168932

The old rubber boot shown in Figure 12.27 has two leaks. To what maximum height can the

Chapter 12 textbook questions

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