Question

A body is projected with a velocity of 30 m/s at an angle of ${30}^{\circ }$ with the vertical. Find the maximum height reached.

Answer 1

Here, $v_o=30m/s{\theta }_o={60}^{\circ }$
So, maximum height, $h_{max}=\frac{{v_o}^{2}si{n}^2{\theta }_o}{2g}=\frac{30\times 30\times si{n}^2{60}^{\circ }}{20}=33.75m$