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 mu…

Question

asked Jul 4, 2020 62.6k views

1 Answer

Sihirbazzz · Answer 1 · 2020-07-08T14:10:14+0000

Answer:

The ball spend
3.5s in the air.

Step-by-step explanation:

The angle which maximize the distance is
$\theta =(\pi)/(4)$ .

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

At
final, the ball will be in the ground and we'll have
y(t_f)=0 .

So,

v_0_yt_f-(1)/(2)gt_f^2=0 ⇒
t_f^2=(2v_0_yt_f)/(g)

⇒
t_f=(2v_0_y)/(g) .

¿But what is the component in
of
,
v_0_y ?

This is
$v_0_y=v_0sin(\theta )=17.18(m)/(s)$ .

So, we finally have:

$t_f=(2v_0sin(\theta ))/(g)=3.5s$ .