Question

Projectile Motion: Other applications


A boy throws a rock with speed y = 18.3 m/s at an angle of ∅ = 57.0

Answer 1

Given:

$v_0=18.3\:\rm m/s$

$\theta=57.0^{\circ}$

The range

$$R=\frac{v_0^2\sin2\theta}{g}=\frac{18.3^2\sin(114^{\circ})}{9.8}=31.2\:\rm m$$

The maximim height

$$h_{\max}=\frac{v_0^2\sin^2\theta}{2g}=\frac{18.3^2\sin^2(57.0^{\circ})}{2*9.8}=12.0\:\rm m$$

The time of flight

$$t=\frac{2v_0\sin\theta}{g}=\frac{2*18.3\sin57^{\circ}}{9.8}=3.13\:\rm s$$