Two stones are projected so as to reach the same horizontal range on a horizontal surface. The maximum height reached by one exceeds the other by an amount equal to half the sum of the heights attained by them. Then,the angle of projection of the stone which attains the smaller height 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:

$A$ ball is thrown up with a certain velocity at an angle $\theta$ to the horizontal. The kinetic energy $KE$ of the ball varies with horizontal displacement $x$ as

State whether the following statements are true or false:
$(a)$ The time of flight of a projectile depends only on the vertical component of the initial velocity.
$(b)$ The acceleration of a projectile at the maximum height is zero.
$(c)$ The maximum range of a projectile depends on the angle of projection.

$A$ 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,

$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then the product of the two times of flight is proportional to