Browse Questions

# Two bodies are projected at angles $\theta$ and $(90^{\large\circ}-\theta)$ with the horizontal at the same speed.The ratio of their maximum heights is

$\begin{array}{1 1}1 : 1\\1 : \tan \theta\\\tan \theta : 1\\\tan^2\theta : 1\end{array}$

Can you answer this question?

Answer : $\tan^2\theta : 1$
$h_{max}=\large\frac{v^2\sin^2\theta}{2g}$
$h'_{max}=\large\frac{v^2\sin^2(90-\theta)}{2g}$
or $h'_{max}=\large\frac{v^2\cos^2\theta}{2g}$
Dividing,
$\large\frac{h_{max}}{h'_{max}}=\frac{\sin^2\theta}{\cos^2\theta}=\frac{\tan^2\theta}{1}$
answered Aug 13, 2014