Question

A baseball leaves the bat at an angle of 37° above the horizontal at a speed of 120 ft/s. How high does it go? How far forward is it at its highest point?

Answer 1

$$h_{\max}=\frac{(v_0\sin\theta)^2}{2g}=\frac{(120\sin37^\circ)^2}{2*32.17}=81\:\rm ft$$

$$L=\frac{v_0^2\sin2\theta}{2g}=\frac{120^2\sin74^\circ}{2*32.17}=215\:\rm ft$$