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A golfer hits his ball with speed vo at an angle 6 above
Chapter 2, Problem 2.16(choose chapter or problem)
A golfer hits his ball with speed \(v_{\mathrm{o}}\) at an angle \(\theta\) above the horizontal ground. Assuming that the angle \(\theta\) is fixed and that air resistance can be neglected, what is the minimum speed \(v_{\mathrm{o}}\) (min) for which the ball will clear a wall of height h, a distance d away? Your solution should get into trouble if the angle \(\theta\) is such that \(\tan \theta<h / d\). Explain. What is \(v_{\mathrm{o}}\) (min) if \(\theta=25^{\circ}, d=50 \mathrm{~m} \text {, and } h=2 \mathrm{~m}\)?
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QUESTION:
A golfer hits his ball with speed \(v_{\mathrm{o}}\) at an angle \(\theta\) above the horizontal ground. Assuming that the angle \(\theta\) is fixed and that air resistance can be neglected, what is the minimum speed \(v_{\mathrm{o}}\) (min) for which the ball will clear a wall of height h, a distance d away? Your solution should get into trouble if the angle \(\theta\) is such that \(\tan \theta<h / d\). Explain. What is \(v_{\mathrm{o}}\) (min) if \(\theta=25^{\circ}, d=50 \mathrm{~m} \text {, and } h=2 \mathrm{~m}\)?
ANSWER:Step 1 of 3
From equations of motion for a projectile motion we have
\(\begin{array}{l}
x(t)=v_{0} t \cos \theta \\
y(t)=-\frac{1}{2} g t^{2}+v_{0} t \sin \theta
\end{array}\)
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