In punting a football, the kicker tries to maximize both the distance of the kick and its "hang time"—the time that the ball is in the air. a kicker gets off a great punt with a hang time of 5.0 s that lands 50 yards from the kicker. what is the speed of the ball as it leaves the kicker’s foot?

v₀ = 26.17 m/s

Explanation:

If

R = Xmax = 50 yards = (50 yards)*(0.9144 m / 1 yard) = 45.72 m

t = 5 s  (Time of Flight)

we can apply the equation:

R = V₀x*t  ⇒   V₀x = R / t = 45.72 m / 5 s   ⇒   V₀x = 9.144 m/s

then we use the equation of Time of Flight:

t = 2*V₀y / g   ⇒    V₀y = g*t / 2 = (9.81 m/s²)*(5 s) / 2  ⇒    V₀y = 24.525 m/s

Finally we apply

v₀ = √(V₀x² + V₀y²)   ⇒   v₀ = √(9.144² + 24.525²) m/s

⇒   v₀ = 26.17 m/s

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