Chapter 4, Problem 1Q

To determine

The angle of the ball thrown to the maximum.

Answer to Problem 1Q

The angle of the ball thrown to the maximum range is $45°$.

Explanation of Solution

Write the expression for maximum range of the projectile motion of the ball.

  $x_{\text{lands}}=\frac{v_0^2\sin \left(2\theta \right)}{g}$        (I)

Here, $v_0$ is the initial velocity of the ball, $\theta$ is the maximum projectile angle, and $g$ is the acceleration due to gravity.

Conclusion:

From the equation (I) the landing distance of the ball is maximum when the quantity $2\theta$ is maximum.

From the trigonometry relation the maximum value for sine function is 1, if $2\theta =90°$, the corresponding angle is,

  $\begin{array}{c}\theta =\frac{90°}{2}\\ =45°\end{array}$

Therefore, the angle of the ball thrown to the maximum range is $45°$.