Question

A football punter wants to kick the ball so that it is in the air for 3.6 s and lands 50 m from where it was kicked. Assume that the ball leaves 1.0 m above the ground.

At what angle should the ball be kicked?
With what initial speed should the ball be kicked?

Answer 1

H = Vy t - 1/2 g t^2 height after time t

Vy t = H + 1/2 g t^2

Vy = H / t + 1/2 g t

Vy = -1 / 3.6 + 4.9 * 3.6 = 17.4 m/s initial vertical speed

Vx = 50 / 3.6 = 13.9 m/s

tan θ = 17.4 / 13.9 = 1.25 θ = 51.4 deg

Ball should be kicked at 51.4 deg

V = (17.4^2 + 13.9^2)^1/2 = 22.3 m/s initial speed

As a check use range formula

R = 22.3^2 sin 102.8 / 9.80 = 49.5 m

That is slightly less than 50m as it should be since its height would be zero after being given the initial speed