$A$ body moves along a circular path of radius $100 \,m$ and completes one revolution in $40 \,sec$. What is the total distance covered at the end of $2 \,min \,20 \,sec$ (in $\pi$)?

$A$ plane is revolving around the Earth with a speed of $100\, km/hr$ at a constant height from the surface of the Earth. The change in the velocity as it travels a half circle is ......... $km/hr$.

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 $t = 2 \, s$.

Assertion: In a circular motion,work done by centripetal force is not always zero.
Reason: If the speed of the particle increases or decreases in circular motion,then the net force acting on the particle does not remain towards the centre.

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is ......... $^o$

$A$ particle moves from the point $(2.0\hat i + 4.0\hat j) \, m$ at $t = 0$ with an initial velocity $(5.0\hat i + 4.0\hat j) \, m/s$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat i + 4.0\hat j) \, m/s^2$. What is the distance of the particle from the origin at time $t = 2 \, s$?