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Chapter 4: Problem 17

Show that the maximum height reached by a projectile is $$ y_{\max }=\left(v_{0} \sin \phi_{0}\right)^{2} / 2 g $$

Short Answer

Expert verified
The maximum height formula \(y_{\max} = \left(v_{0} \sin \phi_{0}\right)^{2} / 2 g\) defines the maximum height reached by a projectile based on initial velocity, launch angle and gravitational force. This formula is derived from the second equation of motion and confirmed by substituting in the time at which the object reaches maximum height.

Step by step solution

01

Understanding given equation

The formula provided, \(y_{\max }=\left(v_{0} \sin \phi_{0}\right)^{2} / 2 g\), is the equation for the maximum height reached by a projectile. Here, \(v_0\) is the initial velocity, \(\phi_0\) is the initial angle at which the object is projected, \(g\) is the acceleration due to gravity, and \(y_{\max}\) is the maximum height reached by the projectile.
02

Applying the equation of motion

The second equation of motion states that R = \(v_{0} t - \frac{1}{2} g t^{2}\) when initial position is zero or where R is the displacement. At maximum height the final velocity (v) becomes zero. Using equation of motion \(v = v_0 - g t\) we find that the time at which object reaches at maximum height (\(t_{\max}\)) is \(t_{\max} = \frac{v_{0} \sin \phi_{0}}{g}\). Putting this into first equation we get \(y_{\max} = v_{0} (\frac{v_0 \sin \phi_0}{g}) - \frac{1}{2} g (\frac{v_0 \sin \phi_0}{g})^{2}\).
03

Simplifying the equation

When we simplify above equation we get \(y_{\max} = \left(v_{0} \sin \phi_{0}\right)^{2} / 2 g\), which matches the given equation.

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Most popular questions from this chapter

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