Juan Soto hits a pitched ball at a 43 degree angle. It just clears the fence, 68 m away. What is the velocity when it left the bat. Assume the fence is the same height as the ball leaving Juan's bat.

Answer 1

Start with the y direction and Δy = v_o t + 1/2 a t², 0 = (v sin 43) t - 9.8/2 t² = t (v sin 43 - 4.9 t). The solution t =0 is simply when the ball is hit. The second solution gives us v sin 43 = 4.9 t and t = 0.1392 v.

In x, v_x is constant and Δx = v_x t, 68 = (v cos 43) t, and t = 68 / (0.7314 v) = 92.98 /v. Equate the two solutions for time: 0.1392 v = 92.98 /v and v² = 92.98/0.1392, v = 26 m/s.