$A$ ball is thrown at a speed of $20 \,m/s$ at an angle of $30^{\circ}$ with the horizontal. The maximum height reached by the ball is (Use $g=10 \,m/s^2$) (in $\,m$)

$A$ projectile is thrown from a point in a horizontal plane such that the horizontal and vertical velocities are $9.8 \; m/s$ and $19.6 \; m/s$ respectively. It will strike the plane after covering a horizontal distance of ........ $m$.

The maximum height attained by a projectile is increased by $10 \%$ by keeping the angle of projection constant. What is the percentage increase in the time of flight (in $\%$)?

The vertical displacement ($y$ in metre) of a projectile in terms of its horizontal displacement ($x$ in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ ms^{-2}$)

In projectile motion,the modulus of the rate of change of velocity is:

$A$ ball $A$ is projected vertically upwards with a certain initial speed $u$. Another ball $B$ of the same mass is projected at an angle of $30^{\circ}$ with the vertical with the same initial speed $u$. At the highest point,the ratio of the potential energy of ball $A$ to that of ball $B$ will be: $(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2})$