Question

A golf ball is launched at an angle of $\mathbf{20}\mathbf{°}$to the horizontal, with a speed of $\mathbf{60}\mathbf{}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}\mathbf{}$and a rotation rate of $\mathbf{90}\mathbf{}\raisebox{1ex}{$\mathbf{\text{rad}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$. Neglecting air drag, determine the number of revolutions the ball makes by the time it reaches maximum height.

Short answer

The number of revolutions the ball makes when at maximum height is$30$

Step-by-step solution

Given

1. Angle of launch is$20°$
2. Horizontal speed is$60\text{m}/\text{s}$
3. Rotational rate is$90\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{sec}$}\right.$

Understanding the concept

Find time to reach the maximum height using the kinematic equation. Using this time, find the angle in radians. The angle is nothing but angular displacement, which can be converted into a number of rotations using the relationship between one rotation and angle traced during one rotation.

Formula:

${\text{v}}_{fy}={\text{v}}_{0y}+\text{at}$

$\theta -{\theta }_0={\omega }_0\text{t}$

Calculate the number of revolutions the ball makes by the time it reaches maximum height

Time to reach maximum height is as follows:

${\text{v}}_{fy}={\text{v}}_{0y}+\text{at}$

At maximum height, the velocity is zero, so

$$\begin{gathered} 0=60\text{sin}20-9.8\text{t} \\ \Rightarrow \text{t}=2.094\text{sec} \end{gathered}$$

Now, the number of revolutions:

$$\begin{gathered} \theta -{\theta }_0={\omega }_{0}t \\ \Rightarrow \theta -{\theta }_0=90\times 2.094 \\ \Rightarrow \theta -{\theta }_0=188\text{rad} \end{gathered}$$

So

$188\times \frac{1\text{rev}}{2\pi }=29.93\text{rev}$

So, the number of revolutions is $30$.