Question

A person can throw a ball at a maximum horizontal range of 120m. How high would the ball go straight up? [6 marks]

Answer 1

$\theta=45^\circ$

$R=120\:\rm m$

$$R=\frac{v_0^2\sin2\theta}{g}=\frac{v_0^2}{g}$$

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

$$h_{\max}=\frac{120\;\rm m}{2}\sin^245^\circ=30\:\rm m$$