Question

Two balls are projected at an angle $\theta$ and $({90}^{\circ }-\theta )$ to the horizontal with same speed.The ratio of their maximum vertical height is

• A. $1:1$
• B. $tan\theta :1$
• C. $1:tan\theta$
• D. $ta{n}^2\theta :1$

Answer 1

The correct option is D $ta{n}^2\theta :1$

The bullets are fired at the same initial speed

Also, maximun height is given by,

$H=\frac{u^2{\sin }^2\theta }{2g}$

For first case when $\theta =\theta$

$H_1=\frac{u^2{\sin }^2\theta }{2g}$

For second case when $\theta =(90-\theta )$

$H_2=\frac{u^2{\sin }^2(90-\theta )}{2g}$

Dividing $H_1$ by $H_2$

$\frac{H_1}{H_2}=\frac{u^2{\sin }^2\theta }{2g}\times \frac{2g}{u^2{\sin }^2(90-\theta )}={\tan }^2\theta$