$A$ 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 $\pi$)?

$A$ stone projected with a velocity $u$ at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity $u$ at an angle $(\frac{\pi}{2} - \theta)$ with the horizontal,it reaches maximum height $H_2$. The relation between the horizontal range $R$ of the projectile,$H_1$ and $H_2$ is

$A$ particle of mass $1 \ kg$ is projected with an initial velocity $10 \ ms^{-1}$ at an angle of projection $45^{\circ}$ with the horizontal. The average torque acting on the projectile,between the time at which it is projected and the time at which it strikes the ground,about the point of projection in $Nm$ is:

$A$ projectile is fired at an angle of $45^o$ with an initial kinetic energy $K$. What is the kinetic energy of the projectile at its maximum height?

The range of a projectile is $100 \ m$. Its kinetic energy will be maximum after covering a horizontal distance of: (in $m$)

$A$ projectile is thrown with some initial velocity at an angle $\alpha$ to the horizontal. Its velocity when it is at the highest point is $(2/5)^{1/2}$ times the velocity when it is at height half of the maximum height. Find the angle of projection $\alpha$ with the horizontal. (in $^{\circ}$)