Two projectiles P₁ and P₂ thrown with speed in the ratio $\sqrt3$ : $\sqrt2$, attain the same height during their motion. If P₂ is thrown at an angle of 60$^\circ$ with the horizontal, the angle of projection of P₁ with horizontal will be :

<p>We know,</p> <p>Maximum height of a projectile $(H) = {{{u^2}{{\sin }^2}\theta } \over {2g}}$</p> <p>Given, Ratio of initial velocity of two projectile</p> <p>${{{u_1}} \over {{u_2}}} = {{\sqrt 3 } \over {\sqrt 2 }}$</p> <p>Both projectile reach the same maximum height.</p> <p>$\therefore$ ${H_1} = {H_2}$</p> <p>$$ \Rightarrow {{u_1^2{{\sin }^2}{\theta _1}} \over {2g}} = {{u_2^2{{\sin }^2}{\theta _2}} \over {2g}}$$</p> <p>$\Rightarrow u_1^2{\sin ^2}{\theta _1} = u_2^2{\sin ^2}{\theta _2}$</p> <p>$$ \Rightarrow {\left( {{{{u_1}} \over {{u_2}}}} \right)^2} = {\left( {{{\sin {\theta _2}} \over {\sin {\theta _1}}}} \right)^2}$$</p> <p>$$ \Rightarrow {\left( {{{\sqrt 3 } \over {\sqrt 2 }}} \right)^2} = {\left( {{{\sin 60^\circ } \over {\sin {\theta _1}}}} \right)^2}$$</p> <p>$\Rightarrow {3 \over 2} = {3 \over {4{{\sin }^2}{\theta _1}}}$</p> <p>$\Rightarrow \sin _{{\theta _1}}^2 = {1 \over 2}$</p> <p>$\Rightarrow \sin {\theta _1} = {1 \over {\sqrt 2 }} = \sin 45^\circ$</p> <p>$\Rightarrow {\theta _1} = 45^\circ$</p>