JEE Kinematics 2

The vertical velocity of a projectile at time $t$ after firing is given by $\space v(t) = v_0\sin{\theta} - gt$ , where $v_0$ is the initial velocity and $\theta$ , the angle of projectile and $g$ , acceleration of gravity.

The time for the projectile to reach the apex is when $v(t_H) = 0 = v_0 \sin{\theta} - gt_H$

$\Rightarrow t_H = \dfrac {v_0 \sin{\theta}}{g}$ .

The time for the projectile to cover the whole range $t_D = 2t_H = \dfrac {2v_0 \sin{\theta}}{g}$ .

The range covered $D = v_0 \cos {\theta} \times t_D = \dfrac {2v_0^2 \sin{\theta}\cos {\theta}}{g} = \dfrac {v_0^2 \sin{2\theta}}{g}$

$\Rightarrow \sin {2\theta} : \sin {4\theta} : \sin {8\theta} = x:2:2\quad \Rightarrow \sin {4\theta} = \sin {8\theta}$

Since $\sin{60^\circ} = \sin{120^\circ}\quad \Rightarrow \theta = 15^\circ$

$\Rightarrow \dfrac {\sin{30^\circ}} {\sin{60^\circ}} = \dfrac {x}{2} = \dfrac {\frac {1} {2} } {\frac {\sqrt{3}}{2}}\quad \Rightarrow x = \dfrac {2}{\sqrt{3}}$

$\Rightarrow \lfloor 10^4x \rfloor = \lfloor \frac {2}{\sqrt{3}} \times 10^4 \rfloor = \boxed{11547}$

Range of Projectile is equal for $45^{\circ} + \alpha \text{ and } 45^{\circ} - \alpha$

$2\theta = 45^{\circ} - \alpha ..... (1)$ $2\theta = 45^{\circ} + \alpha .....(2)$

We get $\theta = 15^{\circ}$

Ratio is : $\frac{2}{\sqrt{3}} : 2 : 2$