The equation of motion of a projectile is $y = 12x - \frac{3}{4}x^2$. The range of the projectile is $..........\,m$.

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $150\, ms^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is

The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is

$A$ particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

The trajectory of a projectile is

$A$ projectile is projected from the ground with an initial velocity $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$. If the acceleration due to gravity $g$ is along the negative $y$-direction,find the maximum displacement in the $x$-direction (horizontal range).