The punter on a football team tries to kick a football so that it stays in the air for a long "hang time." If the ball is kicked with an initial velocity of 24.8 m/s at an angle of 62.5° above the ground, what is the "hang time"?

Answer 1

use
s=ut + 0.5*a*t^2 in vertical direction
s =0 because there is no vertical displacement at the time it reaches the ground
a= -9.81 ; minus because we consider up as the positive direction.
u= 24.8 sin 62.5 ; vertical component of the Velocity
SO
t= 2*u/ -a = (2*24.8 sin 62.5)/9.81
sorry i don't have a calculator with me