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

The correct option is C 30${}^{\circ }$

Given,

$u_A:u_B=1:\sqrt{2}$

$H_A=H_B$

${\theta }_A={45}^0$

${\theta }_B=?$

The maximum height in projectile motion,

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

$\frac{u_A^{2}si{n}^2{\theta }_A}{2g}=\frac{u_B^{2}si{n}^2{\theta }_B}{2g}$

$(\frac{u_A}{u_B}{)}^2=(\frac{sin{\theta }_B}{sin{\theta }_A}{)}^2$

$sin{\theta }_B=sin{\theta }_A.\frac{u_B}{u_A}$

$sin{\theta }_B=sin{45}^0.\frac{1}{\sqrt{2}}=\frac{1}{2}$

${\theta }_B=si{n}^{-1}\left(0.5\right)={30}^0$

The correct option is C.