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