Two particles are projected from the same point with the same speed at angles of projection $30^{\circ }$ and $60^{\circ }$. The ratio of their maximum heights is:

Solution

1. The maximum height $H$ is given by $H=\frac{v^2{\sin }^2\theta }{2g}$.
2. For ${\theta }_1=30^{\circ }$, $$\begin{array}{rl}{H}_1\propto {\sin }^2(30^{\circ })& =(1/2{)}^2\\ & =1/4\end{array}$$.
3. For ${\theta }_2=60^{\circ }$, $$\begin{array}{rl}{H}_2\propto {\sin }^2(60^{\circ })& =(\sqrt{3}/2{)}^2\\ & =3/4\end{array}$$.
4. The ratio $$\begin{array}{rl}{H}_1/H_2& =(1/4)/(3/4)\\ & =1/3\end{array}$$.
5. Hence the answer is (B).