$A$ particle is projected with kinetic energy $K$ at an angle of $60^o$ with the horizontal. The kinetic energy of the particle at the maximum height is . . . . . . .

The maximum height attained by a projectile is increased by $10\,\%$ by increasing its speed of projection,without changing the angle of projection. The percentage increase in the horizontal range will be $...........\,\%$

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

If a stone is to hit a point which is at a horizontal distance $d$ and at a vertical height $h$ above the point from where the stone is launched,what is the initial speed $u$ if the stone is launched at an angle $\theta$?

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 initial speed of a projectile fired from the ground is $u$. At the highest point during its motion,the speed of the projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is: