The maximum height attained by a projectile is increased by $10\,\%$ by increasing its speed of projection,without changing the angle of projection. The percentage increase in the horizontal range will be $...........\,\%$

$A$ body is projected at an angle of $60^{\circ}$ with the horizontal such that the vertical component of its initial velocity is $40 \ m \ s^{-1}$. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

$A$ body is projected from the ground at an angle of $45^{\circ}$ with the horizontal. Its velocity after $2 \, s$ is $20 \, m/s$. The maximum height reached by the body during its motion is $m$. (use $g = 10 \, m/s^2$)

The equations of motion of a projectile are given by $x = 36t$ metre and $2y = 96t - 9.8t^2$ metre. The angle of projection is:

$A$ helicopter is flying horizontally with a speed $v$ at an altitude $h$ and has to drop a food packet for a man on the ground. What is the distance of the helicopter from the man when the food packet is dropped?

The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre) of a projectile is $y = 3x - 0.8x^2$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ m/s^2$). (in $s$)