A football punter wants a punted football to travel 60.0 m in the horizontal direction, and stay in the air for 5.00s. What must be the magnitude and the direction of the initial velocity of the football?

27.28 m/s at 63.90° to the horizontal

Explanation:

Let vi (x) be the intial velocity of the football in the horizontal direction. Let vi (y) be the initial velocity and vf (y) be the final velocity of the football in the vertical direction. Let Δd be the displacement of the football either in the x or y direction and a be the acceleration of the football in that direction. Let downards be the positive direction and g the gravitational acceleration.

Looking at the horizontal motion of the football:

Δd = vi (x) * t + (1/2) * a*t^2

Now in the horizontal direction a = 0 m/s^2, so

Δd = vi (x) * t

vi (x) = Δd/t

vi (x) = 60/5

vi (x) = 12 m/s

Looking at the vertical motion of the football:

vf (y) = vi (y) + a*t

Now in the vertical direction a = g = 9.8 m/s^2

And if we look at the motion until the top of the path of the projectile,

vf (y) = 0 m/s and the projectile will reach the top of its path at half the total travelling time so t = 2.5 s, therefore:

0 = vi (y) + a*t

vi (y) = - a*t

vi (y) = - 9.8 * 2.5

vi (y) = - 24.5 m/s

vi (y) = 24.5 m/s upwards

The resultant initial velocity V is:

V √ (vi (y) ^2 + vi (x) ^2)

V = √ (24.5^2 + 12^2)

V = 27.28 m/s

And the football's initial direction Ф taken from the horizontal is:

tanФ = vi (y) / vi (x)

tanФ = 24.5/12

Ф = 63.90°

Therefore the football's initial velocity and direction is 27.28 m/s at 63.90° to the horizontal.