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 equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal is given as $y = x - \frac{gx^2}{2}$. The initial velocity of the ball is $[\text{Given} : \tan 45^{\circ} = 1, \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

At what horizontal distance does a projectile reach its minimum kinetic energy?

$A$ particle $P$ is projected with velocity $u_1$ at an angle of $30^\circ$ with the horizontal. Another particle $Q$ is projected vertically upwards with velocity $u_2$ from the point directly below the maximum height of $P$. The condition for the collision of both particles is:

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

An object is projected at an angle of $45^\circ$ with the horizontal. The ratio of the horizontal range to the maximum height reached is: