The kinetic energy of a particle moving along a circle of radius $R$ depends on the distance $s$ as $K = as^2$,where $a$ is a constant. Then the force acting on the particle is

$A$ sportsman runs around a circular track of radius $r$ such that he traverses the path $ABAB$. The distance travelled and displacement,respectively,are

$A$ cart moves with a constant speed along a horizontal circular path. From the cart,a particle is thrown up vertically with respect to the cart.

$A$ projectile is fired at a speed of $100 \, m/s$ at an angle of $37^o$ above the horizontal. At the highest point,the projectile breaks into two parts of mass ratio $1:3$,the smaller coming to rest. Then the distance of the heavier part from the launching point is ........... $m$.

$A$ projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \; m/s$. After $10 \; s$,its inclination with the horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \; m/s^2)$

$A$ particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle $\theta$ to the horizontal,the maximum height attained by it equals $4R$. The angle of projection,$\theta$,is then given by