During a baseball game, a batter hits a pop up to a fielder 65m away. The acceleration of gravity is 9.8m/s^2. If the ball remains in the air for 5.7s, how high does it rise? Answer in units of m.

Answer 1

The ball spends half of the 5.7 s going up, and half coming down. So, in the last 2.85 s, the vertical distance it falls is
H = (g/2)(2.85)^2 = 39.8 meters.
That is also the vertical distance that it rises.

The distance of the fielder is not needed. You could use it to obtain the horizontal velocity (11.4 m/s) and launch angle. Vertical initial velocity = g*2.85 s = 27.93 m/s
Launch angle = arctan (27.93/11.4) = 67.8 degrees