Question

Two balls are thrown with the same speed but making complementary angle with the horizontal. If the angle of projection of one is ${60}^{\circ }$ and its maximum height is $h$, then the maximum height of the other will be

• A. $\frac{h}{3}$
• B. $2h$
• C. $4h$
• D. $6h$

Answer 1

The correct option is A $\frac{h}{3}$

Let ${\theta }_1$ and ${\theta }_2$ be the angles of projection of two balls thrown with same speed.
Let $H_1$ and $H_2$ be the maximum height of the projectiles of two balls.
According to question
${\theta }_1+{\theta }_2={90}^{\circ }(\because \text{Angles are complimentary}$
If ${\theta }_1={60}^{\circ },$ then
${\theta }_2={30}^{\circ }$
$H_1=\frac{u^2{\sin }^2{\theta }_1}{2g}=h$
$\Rightarrow \frac{u^2}{2g}\times \frac{3}{4}=h........\text{(i)}$
$H_2=\frac{u^2{\sin }^2{\theta }_2}{2g}=\frac{1}{4}\times \frac{u^2}{2g}.......\text{(ii)}$
Using $\text{(i)}$ and $\text{(ii)}$, we get
$H_2=h/3$

Aliter,
$\frac{H_1}{H_2}=ta{n}^2{\theta }_1$ (we can directly use this result for bodies thrown at same speed at complementory angles)
$\frac{H_1}{H_2}=ta{n}^2{60}^{\circ }$
$\frac{H_1}{H_2}=\frac{3}{1}$