$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The angle of projection of the projectile is:

$A$ particle is projected from the ground with velocity $u$ at an angle $\theta$ with the horizontal. The horizontal range,maximum height,and time of flight are $R, H$,and $T$ respectively. They are given by $R = \frac{u^2 \sin 2\theta}{g}$,$H = \frac{u^2 \sin^2 \theta}{2g}$,and $T = \frac{2u \sin \theta}{g}$. Now,keeping $u$ fixed,$\theta$ is varied from $30^o$ to $60^o$. Then,

$A$ particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momentum of the particle about $P$ at $t = \frac{V_0}{g}$ is:

$A$ projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \; m/s$. After $10 \; s$,its inclination with the horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \; m/s^2)$

$A$ car moving with a certain velocity jumps from an inclined plane placed at one bank of a river and reaches the other bank by attaining a maximum height of $80 \,m$. If the same car, moving with the same velocity, jumps from another inclined plane having a different angle of inclination and reaches the same point on the other bank by attaining a maximum height of $45 \,m$, then the width of the river is: (in $\,m$)

$A$ fighter plane flying horizontally at an altitude of $1.5 \; km$ with a speed of $720 \; km/h$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with a muzzle speed of $600 \; m s^{-1}$ to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take $g = 10 \; m s^{-2}$).