$A$ particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be

Two paper screens $A$ and $B$ are separated by a distance of $100 \,m$. $A$ bullet penetrates $A$ and $B$ at points $P$ and $Q$ respectively,where $Q$ is $10 \,cm$ below $P$. If the bullet is travelling horizontally at the time of hitting $A$,the velocity of the bullet at $A$ is nearly .......... $m/s$.

$A$ projectile is projected with kinetic energy $K$. If it has the maximum possible horizontal range,then its kinetic energy at the highest point will be ......... $K$.

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

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: Two identical balls $A$ and $B$ thrown with the same velocity '$u$' at two different angles with the horizontal attain the same range $R$. If $A$ and $B$ reach maximum heights $h_{1}$ and $h_{2}$ respectively,then $R = 4 \sqrt{h_{1} h_{2}}$.
Reason $R$: The product of the said heights is $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta}{2g}\right)$.
Choose the $CORRECT$ answer.

$A$ projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of the Earth. For the same $v$ and $\theta$,its range on the surface of the Moon will be: