Four bodies $P, Q, R$ and $S$ are projected with equal velocities having angles of projection $15^{\circ}, 30^{\circ}, 45^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The body having the shortest range is

$A$ particle of mass $m$ performs uniform circular motion of radius $r$ with linear speed $v$ under the application of force $F$. If $m$,$v$,and $r$ are all increased by $20 \%$,the necessary change in force required to maintain the particle in uniform circular motion is: (in $\%$)

$A$ body of mass $100 \, g$ is projected at an angle of $30^\circ$ with the horizontal with a velocity of $20 \, m \, s^{-1}$. What is the change in its momentum at the maximum height in $kg \, m \, s^{-1}$?

$A$ point moves in the $xy$-plane according to the following equations: $x = a \sin \omega t$ and $y = a(1 - \cos \omega t)$,where $a$ and $\omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.

$A$ horizontal plane supports a stationary vertical cylinder of radius $R = 1 \ m$ and a disc $A$ attached to the cylinder by a horizontal thread $AB$ of length $l_0 = 2 \ m$ (seen in figure,top view). An initial velocity $v_0 = 1 \ m/s$ is imparted to the disc as shown in the figure. How many seconds will it move along the plane until it strikes against the cylinder? (All surfaces are assumed to be smooth.)

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of $120 \ m$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is.