Hey mate here your answer Explanation:Let US consider that the maximum height covered by the projectile is H and the angle of projection is θ.It is given that,v_{H}=\sqrt{\frac{2}{5}} v_{H / 2}v H = 52 v H/2 So the maximum height of a projectile motion can be found asH=\frac{u^{2} \sin ^{2} \theta}{2 g} \quad \rightarrow(1)H= 2gu 2 sin 2 θ →(1)In the above equation, u is the initial velocity and g is the acceleration DUE to gravity.We know that velocity at maximum height is v_{H}=u \cos \thetav H =ucosθSquaring on both SIDES, we get v_{H}^{2}=u^{2} \cos ^{2} \theta \quad \rightarrow(2)v H2 =u 2 cos 2 θ→(2)From Newton’s second law of motion:v^{2}=u^{2}-2 a s \quad \rightarrow(3)v 2 =u 2 −2as→(3)We can substitute v=v_{H / 2}v=v H/2 , as the velocity at half the maximum height and displacement (s) can be replaced by half the maximum height, i.e., \mathrm{s}=\mathrm{H} / 2\ in\ (3)s=H/2 in (3)v_{H / 2}^{2}=u^{2}-2 g\left(\frac{H}{2}\right) \quad \rightarrow(4)v H/22 =u 2 −2g( 2H )→(4)The H in eqn (4) can be replaced with the value of H in eqn (1)v_{H / 2}^{2}=u^{2}-2 g\left(\frac{\frac{u^{2} \sin ^{2} \theta}{2 g}}{2}\right) \quad \rightarrow(5)v H/22 =u 2 −2g( 22gu 2 sin 2 θ )→(5)v_{H / 2}^{2}=u^{2}-2 g\left(\frac{u^{2} \sin ^{2} \theta}{4 g}\right) \quad \rightarrow(6)v H/22 =u 2 −2g( 4gu 2 sin 2 θ )→(6)It is given that v_{H}=\sqrt{\frac{2}{5}} v_{H / 2}v H = 52 v H/2 So squaring on both sides, we get v_{H}^{2}=\frac{2}{5} v_{H / 2}^{2} \rightarrow(7)v H2 = 52 v H/22 →(7)So, substitute eqn (2) and eqn (6) in eqn (7)u^{2} \cos ^{2} \theta=\frac{2}{5}\left[u^{2}-2 g\left(\frac{u^{2} \sin ^{2} \theta}{4 g}\right)\right]u 2 cos 2 θ= 52 [u 2 −2g( 4gu 2 sin 2 θ )](We know that, \cos ^{2} \theta+\sin ^{2} \theta=1cos 2 θ+sin 2 θ=1 )\therefore u^{2}\left(1-\sin ^{2} \theta\right)=\frac{2 u^{2}}{5}\left[1-2\left(\frac{\sin ^{2} \theta}{4}\right)\right]∴u 2 (1−sin 2 θ)= 52u 2 [1−2( 4sin 2 θ )]\left(1-\sin ^{2} \theta\right)=\frac{2}{5}\left[1-\left(\frac{\sin ^{2} \theta}{2}\right)\right](1−sin 2 θ)= 52 [1−( 2sin 2 θ )]1-\sin ^{2} \theta=\frac{2}{5}-\frac{1}{5} \sin ^{2} \theta1−sin 2 θ= 52 − 51 sin 2 θ\frac{1}{5} \sin ^{2} \theta-\sin ^{2} \theta=\frac{2}{5}-1 51 sin 2 θ−sin 2 θ= 52 −1-\frac{4}{5} \sin ^{2} \theta=-\frac{3}{5}− 54 sin 2 θ=− 53 \sin ^{2} \theta=\frac{3}{5} \times \frac{5}{4}sin 2 θ= 53 × 45 \sin ^{2} \theta=\frac{3}{4}sin 2 θ= 43 \therefore \sin \theta=\frac{\sqrt{3}}{2}∴sinθ= 23 \bold{\therefore \theta=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=60^{\circ}}∴θ=sin −1 ( 23 )=60 ∘ Thus the angle of projection is 60°.