A football is punt 50ft above ground with a hang time of 4.0 s and a horizontal distance of 40 yd.

What is the magnitude and direction of the initial velocity? g= 32.2 ft/s^2.

We know that the ball hits the ground after 4.0 seconds by the definition of hang time. If the initial VERTICAL velocity of the football is v, then the height of the football after time t is:
h(t) = (1/2)at^2 + vt + d₀
= (1/2)(-32.2)t^2 + vt + 50
= -16.1t^2 + vt + 50.

The ball hits the ground after 4 seconds, so h(4) = 0 and:
-16.1(4)^2 + 4v + 50 = 0 ==> v = 51.9 ft/s.

The ball travels a horizontal distance of 40 yards = 120 feet in 4 seconds, so the horizontal velocity of the ball is 120/4 = 30 ft/s. Using the Pythagorean Theorem, the magnitude of the initial velocity is:
||v|| = √(51.9^2 + 30^2) = 60 ft/s,

while the direction is:
θ = arctan(51.9/30) = 60° with respect to the horizontal.

I hope this helps!