$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^\circ$ with the horizontal. When the particle lands on the level ground,the magnitude of the change in its momentum will be

$A$ particle of mass $m$ is projected at a velocity $u$,making an angle $\theta$ with the horizontal ($x$-axis). If the angle of projection $\theta$ is varied keeping all other parameters same,then the magnitude of angular momentum $(L)$ at its maximum height about the point of projection varies with $\theta$ as,

$A$ projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \; m/s$. After $10 \; s$,its inclination with the horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \; m/s^2)$

The position vector of a particle moving in a plane is given by $r = a \cos \omega t \hat{i} + b \sin \omega t \hat{j}$,where $\hat{i}$ and $\hat{j}$ are the unit vectors along the rectangular axes $X$ and $Y$; $a$,$b$,and $\omega$ are constants,and $t$ is time. The acceleration of the particle is directed along the vector:

The velocity of a body at time $t = 0$ is $10\sqrt{2} \, m/s$ in the north-east direction and it is moving with an acceleration of $2 \, m/s^2$ directed towards the south. The magnitude and direction of the velocity of the body after $5 \, s$ will be

$A$ particle is projected from the ground with velocity $u$ at an angle $\theta$ with the horizontal. The horizontal range,maximum height,and time of flight are $R, H$,and $T$ respectively. They are given by $R = \frac{u^2 \sin 2\theta}{g}$,$H = \frac{u^2 \sin^2 \theta}{2g}$,and $T = \frac{2u \sin \theta}{g}$. Now,keeping $u$ fixed,$\theta$ is varied from $30^o$ to $60^o$. Then,