Question

A golfer hits a golf ball at an angle of 25 degrees to the ground at a speed of 76 m/s. If the gold ball covers a horizontal distance of 301.5 m:
What is the initial vertical velocity?
What is the initial horizontal velocity?
How long was it in the air?
What is the ball’s maximum height?

Answer 1

The ball's horizontal position x and vertical position y at time t are given by

x = (76 m/s) cos(25º) t

y = (76 m/s) sin(25º) t - 1/2 g t²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball's initial vertical velocity is (76 m/s) sin(25º) ≈ 32.12 m/s.

Its initial horizontal velocity is (76 m/s) cos(25º) ≈ 68.88 m/s.

The ball stays in the air for as long as y > 0. Solve y = 0 for t :

(76 m/s) sin(25º) t - 1/2 g t² = 0

t ((76 m/s) sin(25º) - 1/2 g t ) = 0

t = 0   or   (76 m/s) sin(25º) - 1/2 g t = 0

Ignore the first solution.

(76 m/s) sin(25º) - 1/2 g t = 0

(76 m/s) sin(25º) = (4.90 m/s²) t

t = (76 m/s) sin(25º) / (4.90 m/s²)

t ≈ 6.55 s

Recall that

v² - u² = 2 a ∆y

where u and v denote initial and final velocities, a is acceleration, and ∆y is displacement. At maximum height, the ball has zero vertical velocity, and taking the ball's starting position on the ground to be the origin, ∆y refers to the maximum height. So we have

0² - ((76 m/s) sin(25º))² = 2 (-g) ∆y

∆y = ((76 m/s) sin(25º))² / (2g)

∆y ≈ 52.6 m