Question

A golfer hits the ball off the tee at an angle of thirty-five degrees from the horizontal with a speed of 46 m/s. It lands on the green, which is elevated 5.50 m higher than the tee. How much time elapsed from when the ball was hit to when it landed on the green?

Answer 1

Answer:

t = 5.16 seconds.

Explanation:

The flight time can be found using the following equation:

$y_{f} = y_{0} +v_{0y}*t - \frac{1}{2}gt^{2}$

Where:

t: is the time

g: is the gravity = 9.81 m/s²

y₀: is the initial height = 0

$y_{f}$: is the final height = 5.50 m

$v_{0y}$: is the initial vertical velocity = v*sin(35)

v: is the speed = 46 m/s

$5.50 = 46*sin(35)*t - \frac{1}{2}9.81*t^{2}$

By solving the above cuadratic equation we have:

t₁= 0.22 s and t₂= 5.16 s      

We will take the solution equal to 5.16 seconds, since in 0.22 seconds (very short time) the ball is going up and in 5.16 seconds it landed on the green.                        

 

Therefore, 5.16 seconds have passed since the ball was hit until it landed.

I hope it helps you!