Calculate Initial Speed of a Football Kicked at 45 Degrees

We can start by using the kinematic equation for projectile motion:

Δx = v₀cos(θ)t

where Δx is the horizontal displacement, v₀ is the initial speed, θ is the angle of launch, and t is the time of flight.

We know that Δx = 23.3 m and θ = 45.0º. We also know that the time of flight is twice the time it takes for the ball to reach its maximum height, which we can find using the vertical kinematic equation:

Δy = v₀sin(θ)t - 1/2gt²

where Δy is the maximum height reached by the ball, and g is the acceleration due to gravity (9.81 m/s²).

Since the ball starts and ends at ground level, we know that Δy = 0. Solving for t, we get:

t = 2v₀sin(θ)/g

Substituting this into the horizontal kinematic equation, we get:

Δx = v₀cos(θ)(2v₀sin(θ)/g)

Simplifying, we get:

Δx = 2v₀²sin(θ)cos(θ)/g

Substituting in our values, we get:

23.3 m = 2v₀²sin(45.0º)cos(45.0º)/9.81 m/s²

Solving for v₀, we get:

v₀ = √(23.3 m * 9.81 m/s² / (2 * sin(45.0º)cos(45.0º)))

v₀ = 29.8 m/s

Therefore, the initial speed of the ball was 29.8 m/s.