Answer to Question #267231 in Physics for Aslabanga

Question

A body is projected at an angle of 30° to the horizontal with a velocity of 150ms^{-1 .} calculate the greatest height (g =10)

Answer 1

Given:

$v_0=150\:\rm m/s$

$\theta=30^{\circ}$

$g=10\:\rm m/s^2$

The maximum height

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

$$h_{\max}=\frac{150^2\sin^2 30^{\circ}}{2*10}=281\:\rm m$$