$A$ 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 :

The velocity of a particle moving in a curvilinear path in a horizontal $X-Y$ plane varies with time as $\vec{v} = (2t\hat{i} + t^2\hat{j}) \ m/s$. Here,$t$ is in seconds. At $t = 1 \ s$:

$A$ particle starts from the origin at $t=0$ with an initial velocity of $3.0 \hat{i} \; m/s$ and moves in the $x-y$ plane with a constant acceleration $(6.0 \hat{i} + 4.0 \hat{j}) \; m/s^2$. The $x$-coordinate of the particle at the instant when its $y$-coordinate is $32 \; m$ is $D$ meters. The value of $D$ is

$A$ projectile is thrown with velocity $v$ making an angle $\theta$ with the vertical. It reaches a maximum height $H$. The time of flight is:

The ratio between the maximum range and the square of the time of flight in projectile motion is:

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