$A$ projectile fired with initial velocity $u$ at an angle $\theta$ has a range $R$. If the initial velocity is doubled at the same angle of projection,then the new range will be:

An aeroplane is flying horizontally with a velocity of $600 \, km/h$ at a height of $1960 \, m$. When it is vertically above a point $A$ on the ground,a bomb is released from it. The bomb strikes the ground at point $B$. The distance $AB$ is

$A$ gun and a target are at the same horizontal level separated by a distance of $600 \,m$. The bullet is fired from the gun with a velocity of $500 \,ms^{-1}$. In order to hit the target, the gun should be aimed to a height $h$ above the target. The value of $h$ is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,m$)

Four objects $P, Q, R,$ and $S$ are projected with the same velocity at angles of $15^o, 30^o, 45^o,$ and $60^o$ respectively. Which of these will have the minimum range?

The trajectory of a projectile projected from the origin is given by the equation $y = x - \frac{2x^2}{5}$. The initial velocity of the projectile is:

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 $\%$)?