Question

a baseball is thrown to a distance of 890 m at an angle of 30 degree above horizontal. what is the initial speed of the baseball

Answer 1

Given:

$R=890\;\rm m$

$\theta=30^\circ$

The range of projectile is given by

$$R=\frac{v_0^2\sin 2\theta}{g}$$

Hence, the initial velocity

$$v_0=\sqrt{\frac{gR}{\sin 2\theta}}=\sqrt{\frac{9.8*890}{\sin 60^\circ}}=100\:\rm m/s$$