You want to throw a baseball over a wall that has a height of 10.0 m (relative to your release point) and is6.00 m away from you. Assume that the wall is negligibly thick and the baseball is small enough that youconsider it as a point particle. Ignore air resistance. What is the minimum initial speed that the ball can have and still clear the wall?

v₀ = 14.6 m / s

Explanation:

This is a projectile launch exercise, in this case we must stop the wall, we can assume that at this point the ball is at its maximum height

as the height is maximum its vertical speed is zero

         v_{y} = - g t

          y = v_{oy} t - ½ g t²

the horizontal distance is

          x = v₀ₓ t

we use trigonometry to find the components of the velocity

          sin θ = v_{oy} / v₀

          cos θ = v₀ₓ / v₀

          v₀ₓ = v₀ cos θ

          v_{oy} = v_{o} sin θ

we substitute the values

           0 = v₀ sin θ - g t

         10 = v₀ sin θ t - ½ g t²

           6 = v₀ cos θ t

we have three equations and three unknowns by which the system can be solved

let's use the first and second equations,

       10 = (gt) t - ½ g t²

       10 = g t² - ½ g t²

       10 / 9.8 = t² (1- ½)

        1.02 = ½ t²

         t = √ 2.04

         t = 1,429 s

this is the time to hit the wall

now let's use the first and third equations, we substitute the data

          v₀ sin θ = g t

          v₀ cos θ = x / t

we divide and simplify

           tan θ = g t² / x

           θ = tan⁻¹ g t² / x

           θ = tan⁻¹ (9.8 1.429²/6)

           θ = 73.3°

we substitute in the third equation

          x = v₀ cos 73.3   t

          v₀ = x / t cos 73.3

          v₀ = 6 / (1.429 cos 73.3)

          v₀ = 14.6 m / s

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