The equation of a projectile is $y = 16x - \frac{5x^2}{4}$. The horizontal range is .......... $m$.

$A$ bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is (in $^{\circ}$)

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]

The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre) of a projectile is $y = 3x - 0.8x^2$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ m/s^2$). (in $s$)

If the equation of motion of a projectile is given by $y = 12x - \frac{3}{4}x^2$ and its horizontal component of velocity is $3 \ m/s$,then find its range. $(g = 10 \ m/s^2)$ (in $m$)

The angle of projection of a projectile for which its initial kinetic energy becomes half at its maximum height is (in $^{\circ}$)