the speed of a projectile when it is at its greatest height is root(2/5) times its speed at half the - Physics - Motion In A Plane

Question

the speed of a projectile when it is at its greatest
height is root(2/5) times its speed at half the max height what is
the angle of projection ?

Himanshu · Accepted Answer

Time taken to reach the maximum height = $\frac{u \sin θ}{g}$

Maximum height = $\frac{u^{2} \sin^{2} θ}{2 g}$

Half the maximum height = $\frac{u^{2} \sin^{2} θ}{4 g}$

Horizontal velocity at half the maximum height = $u \cos θ$

Vertical velocity at half the maximum height = $\sqrt{u^{2} \sin^{2} θ - 2 g (\frac{u \sin θ}{2 g})} = \frac{u \sin θ}{\sqrt{2}}$

Velocity at half the maximum height = $\sqrt{u^{2} \cos^{2} θ + \frac{u^{2} \sin^{2} θ}{2}}$

$u \cos θ = \sqrt{\frac{2}{5}} \sqrt{u^{2} \cos^{2} θ + \frac{u^{2} \sin^{2} θ}{2}} \Rightarrow u^{2} \cos^{2} θ = \frac{2}{5} u^{2} (1 - \sin^{2} θ + \frac{\sin^{2} θ}{2}) \Rightarrow 5 (1 - \sin^{2} θ) = 2 - \sin^{2} θ \Rightarrow 3 = 4 \sin^{2} θ \Rightarrow \sin θ = \frac{\sqrt{3}}{2} \Rightarrow θ = \frac{π}{3}$

the speed of a projectile when it is at its greatest height is root(2/5) times its speed at half the max height . what is the angle of projection ?