In 1926 Babe Ruth hit a home run that was estimated to have landed 180. meters from home plate and to have reached a height of 21.3 meters. What was the initial velocity of the ball?

v = 47.85 m / s ,   θ = 64.7º

Explanation:

This is a missile throwing exercise.

Let's find the speed to reach the maximum height, at this point the vertical speed is zero

        = v_{oy}^{2} - 2 g y

         0 = v_{oy}^{2} - 2gy

        v_{oy} = √2gy

let's calculate

        v_{oy} = √ (2 9.8 21.3)

        v_{oy} = 20.43 m / s

now we can calculate the time it takes to get to this point

      vy = v_{oy} - g t

       t = v_{oy} / g

       t = 20.43 / 9.8

       t = 2.08 s

in projectile launching, the time it takes for the body to rise is the same as the time it takes to go down, so the total launch time is

      = 2 t

     t_{v} = 2 2.08 = 4.16 s

let's use the horizontal throw ratio

      x = v₀ₓ t_{v}

      v₀ₓ = x / t_{v}

      v₀ₓ = 180 / 4.16

      v₀ₓ = 43.27 m / s

initial velocity is

      v = √ (v₀ₓ² + v_{oy}^{2})

      v = √ (20.43² + 43.27²)

      v = 47.85 m / s

with an angle of

      tan θ = I go / vox

      θ = tan⁻¹ (43.27 / 20.43)

      θ = 64.7º

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