h = Vy t - 1/2 g t^2 = -1.07 m final height of ball relative to being hit
Vy = 37 * sin 53.1 = 29.6 m/s initial vertical velocity
h = 29.6 t - 9.8 /2 t^2
4.9 t^2 -29.6 t - 1.07 = 0
Solving the quadratic for t = 6.08 sec for time of flight
Vy - g t = 0 where t is time to reach max height
t = 29.6 / 9.8 = 3.02 sec to reach max height
S = Vx * t where t is the distance traveled
S = V0 cos 53.1 * 6.08 = 37 * .60 * 6.08 = 135 m about 443 ft