Chapter 3, Problem 33P

To determine

The angle for which the range of projection equal its maximum height.

Answer to Problem 33P

The horizontal range and maximum height are equal for angle 75.96°.

Explanation of Solution

Given:

The horizontal range and maximum height is equal.

Formula used:

Height and range are:

  $\begin{array}{l}H=\frac{u^2{\sin }^2\theta }{2g}\\ R=\frac{u^2\sin 2\theta }{g}\end{array}$

Calculation:

Let $\text{θ}$ be the angle of projection for which horizontal range and maximum height are equal.

  $\begin{array}{l}H=R\\ \frac{u^2{\sin }^2\theta }{2g}=\frac{u^2\sin 2\theta }{g}\\ \frac{u^2{\sin }^2\theta }{2g}=\frac{u^{2}2\sin \theta \cdot \cos \theta }{g}\\ \tan \theta =4\\ \theta ={\tan }^{-1}\left[4\right]=75.96°\end{array}$

Conclusion:

Thus, the horizontal range and maximum height are equal for angle 75.96°.