A player hits a baseball with an initial velocity of 117 ft/sec at an elevation angle of 50°. When struck, the baseball is 2 feet above the ground. What is the maximum height the baseball reaches?

The ball’s vertical launch speed is 117 sin(50°) ≈ 89.6 ft/sec. Its rise above the launch point is vᵧ²/(2g) ≈ 89.6²/(2 × 32) ≈ 125 feet. Adding the 2 feet above the ground where it was hit gives about 127 feet, so the closest choice is 126 feet.

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