Question

Two projectiles $A$ and $B$ thrown with speeds in the ratio $1: \sqrt{2}$ acquired the same heights. If A is thrown at an angle of $45^{\circ}$ with the horizontal, the angle of projection of $B$ will be:

• A. $ 0^{\circ} $
• B. $ 60^{\circ} $
• C. $ 30^{\circ} $
• D. $45^{\circ} $

Answer 1

Answer: (C)

$H =\frac{ v ^{2} \sin ^{2} \alpha}{2 g }$
where $v =$ speed and $\alpha=$ angle
Now, both $A$ and $B$ have same height.
$\Rightarrow v _{ a }^{2} \frac{\sin ^{2} 45^{\circ}}{2 g }= v _{ b }^{2} \frac{\sin ^{2} \phi}{2 g }$
$\Rightarrow \sin ^{2} \phi=\left(\frac{ v _{ a }^{2}}{ v _{ b }^{2}}\right) \sin ^{2} 45=\frac{1}{4}$
$\Rightarrow \sin \phi=\frac{1}{2}$
$\Rightarrow \phi=30^{\circ}$