The trajectory of a projectile,projected from the ground,is given by $y = x - \frac{x^2}{20}$,where $x$ and $y$ are measured in meters. The maximum height attained by the projectile will be $...........\,m$.

$A$ particle is projected with speed $u$ at an angle $\theta$ with the horizontal from the ground. If it is at the same height from the ground at times $t_1$ and $t_2$,then its average velocity in the time interval $t_1$ to $t_2$ is .........

The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is

If the initial velocity in the horizontal direction of a projectile is the unit vector $\hat{i}$ and the equation of the trajectory is $y = 5x(1 - x)$,find the $y$-component vector of the initial velocity. (Take $g = 10\,m/s^2$) (in $,\hat{j}$)

$A$ ball is projected with velocity $V_0$ at an angle of elevation $30^o$. Mark the correct statement.

$A$ particle $P$ is projected with velocity $u_1$ at an angle of $30^\circ$ with the horizontal. Another particle $Q$ is projected vertically upwards with velocity $u_2$ from the point directly below the maximum height of $P$. The condition for the collision of both particles is: