When a body moves with a constant speed along a circle,

$A$ projectile of mass $m$ is fired with velocity $v$ from the point $P$ at an angle $45^{\circ}$ with the horizontal. The magnitude of change in momentum,when it passes through the point $Q$ on the same horizontal line on which $P$ lies,is :

$A$ ball is thrown up with a certain velocity at an angle $\theta$ to the horizontal. The kinetic energy $KE$ of the ball varies with horizontal displacement $x$ as

$A$ particle moves along a parabolic path $y = 9x^2$ in such a way that the $x$-component of velocity remains constant and has a value of $\frac{1}{3} \; m/s$. The acceleration of the particle is $....... \; m/s^2$.

$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

The coordinates of a moving particle at any time $t$ are given by $x = \alpha t^3$ and $y = \beta t^3$,where $\alpha$ and $\beta$ are constants. The speed of the particle at time $t$ is given by: