Question

Two projectiles \( P_{1} \) and \( P_{2} \) thrown with speed in the ratio \( \sqrt{3}: \sqrt{2} \), attain the same height during their motion. If \( P _{2} \) is thrown at an angle of \( 60^{\circ} \) with the horizontal, the angle of projection of \( P_{1} \) with horizontal will be : A \( 15^{\circ} \) B \( 30^{\circ} \) C \( 45^{\circ} \) D \( 60^{\circ} \) Answer Given By Candidate:C

Comments

On solving I'm getting sin^2 x = 1/√3

Answer 1

By the equation of height of projectile motion,we can solve this

Answer 2

Let the angle of projection of P1 be A and the velocities of projection of P1 and P2 be √3u and √2u respectively then maximum height attained  by each will be H

So H= (√3usinA)^2/2g= (√2usin60)^2/2g

Hence sinA=1/√2

A= 45^o