Question

Two projectiles are projected at angles $\left(\theta \right)$ and $(\frac{\pi }{2}-\theta )$ to the horizontal respectively with same speed $20\text{m/s}$. One of them rises $10\text{m}$ higher than the other. Find the angle of projection $\theta$.

• A.
  ${30}^{\circ }$
• B.
  ${45}^{\circ }$
• C.
  ${37}^{\circ }$
• D.
  ${53}^{\circ }$

Answer 1

The correct option is A

${30}^{\circ }$
Maximum height for a projectile is given by the formula $$H=\frac{u^2{\sin }^2\theta }{2g}$$ Let, $h_1$ be the maximum height for $1^{st}$ projectile. So,

$h_1=\frac{u^2{\sin }^2\theta }{2g}$

Given, $u=20\text{m/s};g=10{\text{m/s}}^2$

$\Rightarrow h_1=\frac{{20}^2\times {\sin }^2\theta }{2\times 10}$

$\Rightarrow h_1=20{\sin }^2\theta$

Let $h_2$ be the maximum height for second projectile. So,

$h_2=\frac{u^2{\sin }^2(\frac{\pi }{2}-\theta )}{2g}$

Substituting the values,

$h_2=\frac{{20}^2\times {\sin }^2(\frac{\pi }{2}-\theta )}{2\times 10}$

$\Rightarrow h_2=20{\cos }^2\theta$

Given that,

$h_2-h_1=10\text{m}$

$\Rightarrow 20{\cos }^2\theta -20{\sin }^2\theta =10$

$\Rightarrow {\cos }^2\theta -{\sin }^2\theta =1/2$

$\Rightarrow \cos 2\theta =\frac{1}{2}$

$\Rightarrow 2\theta ={\cos }^{-1}\frac{1}{2}$

$\Rightarrow 2\theta ={60}^{\circ }$

$\Rightarrow \theta ={30}^{\circ }$

Hence, option $\left(a\right)$ is the correct choice.