Kinematics problem from JEE advanced

Initial speed of the projectile = U.; Horizontal component = Ux, Vertical compnent = Uy

Angle of inclination with the horizontal = angle of projection = θ Ux = U cos θ and Uy = U sin θ => U² = (Ux)² + (Uy)²

When it is at any height h : Speed = V ; Horizontal component = Vx, Vertical component = Vy Vx = Ux , => (Vx)² = (Ux)² ; (Vy)² = (Uy)² - 2gh

V² = (Vx)² + (Vy)² = (Ux)² + (Uy)² - 2gh = U² - 2gh ......... (1)

At maximum height : h = H and V = V1 , => (V1)² = U² - 2g H ...... (2)

At half the maximum height : h = H/2 and V = V2, => (V2)² = U² - gH ........ (3)

As per the question, (V1)² / (V2)² = (U² - 2gH) / (U² - gH) = (2 / 5) => 5U² - 10gH = 2U² - 2gH => 3U² = 8gH => U² = (8gH / 3) ........ (4)

Again, at maximum height, (V1)x = Ux = U cos θ and (V1)y = 0 => (V1)² = U² cos² θ = U² - 2gH [ from (2) ] => U² ( 1 - cos² θ ) = 2gH = U² sin² θ = 2gH ......... (5)

Dividing (5) by (4) , sin² θ = (2*3) / 8 = 6/8 = 3/4 => sin θ = √3 / 2 , => θ = 60 deg

The vectorial equation of the particle is $r(t)=v_0\cos \theta \textbf{i}+v_0\sin \theta t-\frac{1}{2}gt^2\textbf{j},$ the speed is $||r'(t)||$ at time $t$ .\ So the greatest height comes when $t=\frac{v_0 \sin \theta}{g},$ let's denote the greatest high with $t_{max}$ $r(t_{max})=\frac{v_0^2\sin \theta \cos \theta}{g}\textbf{i}+\frac{v_0^2\sin^2 \theta }{2g}\textbf{j},$ so the half of the greatest heigh is $HalfHeight=\frac{v_0^2\sin^2 \theta }{4g},$ we just need to find when $r(t)=HalfHeight$ and that occurs when \ $t_{maxhalf}=(1-\frac{\sqrt{2}}{2})\frac{v_0\sin \theta }{g}$ . Now we can replace the value of $t_{maxhalf}$ in the speed $||r'(t_{maxhalf})||=v_0\sqrt{\frac{1+\cos^2 \theta}{2}},$ but the speed of the projectile when it is at its greatest height is $\sqrt{\frac{2}{5}}$ times its speed at half of the maximum height. So we get the equation $v_0\cos \theta=\sqrt{\frac{2}{5}}v_0\sqrt{\frac{1+\cos^2 \theta}{2}}$

$\cos \theta=\sqrt{\frac{1+\cos^2 \theta}{5}}$

$\cos^2\theta=\frac{1+\cos^2 \theta}{5}$ $5\cos^2\theta=1+\cos^2 \theta$ $\cos^2\theta=\frac{1}{4}$ $\cos\theta=\frac{1}{2}$ so $\boxed{\theta=60^\circ}$

In this question, if the factor is $\frac{\sqrt{2}}{5}$ , then On solving , we get $\sin \theta$ = $\frac{\sqrt{46}}{7}$ . If the multiplying factor is $\sqrt{2/5}$ , then we get value of $\theta$ as 60.