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)

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