An object is projected with kinetic energy $K$ from a point $A$ at an angle $60^{\circ}$ with the horizontal. In the absence of air friction, what is the ratio of the difference in kinetic energies at points $B$ and $C$ to the kinetic energy at point $A$ (see figure)?

The coordinates of a particle moving in a plane are given by $x = a \cos(pt)$ and $y = b \sin(pt)$,where $a, b (b < a)$ and $p$ are positive constants of appropriate dimensions. Then:

Assertion $(A)$: When a vehicle takes a turn on the road,it travels along a curved path. Reason $(R)$: In a curved path,the velocity of the vehicle remains the same.

At $t = 0$,a projectile is fired from a point $O$ (taken as origin) on the ground with a speed of $50 \, m/s$ at an angle of $53^{\circ}$ with the horizontal. It passes through two points $A$ and $B$,each at a height of $75 \, m$ above the horizontal,as shown in the figure. Find the distance (in meters) of the particle from the origin at $t = 2 \, s$.

Two bodies are thrown up at angles of $45^\circ$ and $60^\circ$,respectively,with the horizontal. If both bodies attain the same vertical height,then the ratio of the velocities with which these are thrown is:

$A$ 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.