Question

A golfer imparts a speed of 24.3 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. How much time does the ball spend in the air?

Answer 1

Answer:

The ball spend $3.5s$ in the air.

Explanation:

The angle which maximize the distance is $\theta =\frac{\pi}{4}$.

From Kinematics, the position in the component $y$ as a function of time is: $y(t)=v_0_yt-\frac{1}{2}gt^2$, with $g=9.8\frac{m}{s^2}$.

At $t$ final, the ball will be in the ground and we'll have $y(t_f)=0$.

So,

$v_0_yt_f-\frac{1}{2}gt_f^2=0$ ⇒ $t_f^2=\frac{2v_0_yt_f}{g}$

⇒ $t_f=\frac{2v_0_y}{g}$.

¿But what is the component in $y$ of $v$, $v_0_y$?

This is $v_0_y=v_0sin(\theta )=17.18\frac{m}{s}$.

So, we finally have:

$t_f=\frac{2v_0sin(\theta )}{g}=3.5s$.