$A$ person can throw a ball up to a maximum range of $100 \, m$. How high above the ground can he throw the same ball (in $, m$)?

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is:

$A$ 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 is thrown at an angle of $30^{\circ}$ to the horizontal with a velocity of $30\; m/s$. After $1\; s$,its velocity will be (in $m/s$) $\left(g=10\; m/s^{2}\right)$

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways, then the product $t_1 t_2$ is proportional to

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is .............