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 $[\text{Given} : \tan 45^{\circ} = 1, \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

The equation for the trajectory of a projectile is $y = \left( \frac{x}{\sqrt{3}} - \frac{x^2}{60} \right) \text{ m}$. The velocity of projection of the projectile is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

$A$ particle is projected from a horizontal plane such that its velocity vector at time $t$ is given by $\vec{v} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The velocity of projection is........ $ms^{-1}$.

$A$ small object is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity $v_0$. The velocity is averaged for the first $\sqrt{2} \,s$ and the magnitude of the average velocity comes out to be the same as that of the initial velocity, i.e.,$|v_0|$. The magnitude $|v_0|$ will be (take $g=10 \,m/s^2$):

$A$ ball at point '$O$' is at a horizontal distance of $7 \ m$ from a wall. On the wall,a target is set at point '$C$'. If the ball is thrown from '$O$' at an angle $37^{\circ}$ with the horizontal,aiming at the target '$C$',but it hits the wall at point '$D$',which is at a vertical distance '$y_0$' below '$C$'. If the initial velocity of the ball is $15 \ m/s$,find $y_0$ (given $\cos 37^{\circ} = \frac{4}{5}$). (in $m$)