Question

A particle is projected with velocity $u$ at angle ${\theta }_1$ with horizontal. The ratio of range and maximum height is $4$.
When it is projected at angle ${\theta }_2$ with horizontal with same speed, the ratio of range and maximum height is $2$.
Then $\frac{\tan {\theta }_1}{\tan {\theta }_2}$ will be equal to

• A. $4$
• B. $2$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{2}$

We know that $R=\frac{u^2\sin 2\theta }{g}$,
$H=\frac{u^2{\sin }^2\theta }{2g}$
Dividing above equations
$\frac{R}{H}=\frac{4}{\tan \theta }$

Now, $\frac{4}{\tan {\theta }_1}=4$
$\frac{4}{\tan {\theta }_2}=2$
$\Rightarrow \frac{\tan {\theta }_1}{\tan {\theta }_2}=\frac{1}{2}$