The ball's horizontal position <em>x</em> and vertical position <em>y</em> at time <em>t</em> are given by
<em>x</em> = (76 m/s) cos(25º) <em>t</em>
<em>y</em> = (76 m/s) sin(25º) <em>t</em> - 1/2 <em>g</em> <em>t</em>²
where <em>g</em> = 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 <em>y</em> > 0. Solve <em>y</em> = 0 for <em>t</em> :
(76 m/s) sin(25º) <em>t</em> - 1/2 <em>g</em> <em>t</em>² = 0
<em>t</em> ((76 m/s) sin(25º) - 1/2 <em>g</em> <em>t </em>) = 0
<em>t</em> = 0 or (76 m/s) sin(25º) - 1/2 <em>g</em> <em>t</em> = 0
Ignore the first solution.
(76 m/s) sin(25º) - 1/2 <em>g</em> <em>t</em> = 0
(76 m/s) sin(25º) = (4.90 m/s²) <em>t</em>
<em>t</em> = (76 m/s) sin(25º) / (4.90 m/s²)
<em>t</em> ≈ 6.55 s
Recall that
<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>y</em>
where <em>u</em> and <em>v</em> denote initial and final velocities, <em>a</em> is acceleration, and ∆<em>y</em> 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, ∆<em>y</em> refers to the maximum height. So we have
0² - ((76 m/s) sin(25º))² = 2 (-<em>g</em>) ∆<em>y</em>
∆<em>y</em> = ((76 m/s) sin(25º))² / (2<em>g</em>)
∆<em>y</em> ≈ 52.6 m
