A golfer imparts a speed of 30.3 m/s to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green?

Solutions

VerifiedThe initial velocity of the ball is:

$v_0=30.3\,\text{m/s}$

We have to calculate: a) the elapsed time of the ball after it reaches the maximum horizontal distance b) the longest horizontal distance of the ball

$\textbf{(a)}$ The ball will travel the maximum distance if the it is launched at an angle of $\theta=45\text{\textdegree}$, to find the time of flight we can use the following equation in the vertical plane,

$y=v_{0y}t+\dfrac{1}{2}at^2$

where $y$ is the displacement when the ball lands on the green, so it equals zero, since the ball starts and ends at the same elevation. And the initial velocity in the vertical component is given with $v_{0y}=v_{0}\sin(\theta)$, hence:

$0=v_{0}\sin(\theta)t+\dfrac{1}{2}at^2$

$\dfrac{1}{2}at=-v_{0}\sin(\theta)$

$t=-\dfrac{2v_{0}\sin(\theta)}{a}$

substitute with the given values to get:

$\begin{align*} t&=-\dfrac{2(30 \mathrm{~m \cdot s^{-1}})\sin(45)}{(-9.8 \mathrm{~m \cdot s^{-2}})}\\ &=4.33 \mathrm{~s} \end{align*}$

$\boxed{t=4.33 \mathrm{~s}}$