$A$ ball is projected horizontally with a velocity of $5 \text{ m/s}$ from the top of a building $19.6 \text{ m}$ high. How long will the ball take to hit the ground?

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is .......... $^o$

$A$ shell fired from the base of a mountain just clears it. If $\alpha$ is the angle of projection,then the angular elevation of the summit $\beta$ is

$A$ particle of mass $m$ is thrown with an initial speed $u$ at an angle $\theta$ with the horizontal. Find the torque of its weight about the point of projection when it reaches the highest point.

$A$ ground-to-ground projectile is at point $A$ at $t = \frac{T}{3}$,is at point $B$ at $t = \frac{5T}{6}$,and reaches the ground at $t = T$. The difference in heights between points $A$ and $B$ is

$A$ body of mass $10 \ kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20, 10)$. If $T$ is the time of flight,then its momentum vector,at time $t = \frac{T}{\sqrt{2}}$,is. [Take $g = 10 \ m/s^2$]