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 $\text{ m s}^{-1}$)

$A$ 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 $m$)

$A$ ball $P$ is projected at an angle of $60^{\circ}$ with the vertical with a certain initial speed $u$. Another ball $Q$ of the same mass $m$ as that of ball $P$ is projected vertically upwards with the same initial speed $u$. At the highest point, the ratio of the potential energy of ball $P$ to that of ball $Q$ is $(\sin 30^{\circ} = 0.5)$.

$A$ particle is projected with velocity $u$ at an angle $\theta$ with the horizontal. What is the change in its velocity at the maximum height?

$A$ projectile is thrown into space so as to have a maximum possible horizontal range of $400 \, m$. Taking the point of projection as the origin,the coordinates of the point where the velocity of the projectile is minimum are

Which of the following remains constant for a projectile fired from the earth?