$A$ projectile is thrown with velocity $u$ at an angle of $15^o$ and has a range $R$. What will be the range of the projectile if it is thrown with velocity $2u$ at an angle of $45^o$ (in $, R$)?

$A$ projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$ with velocity $v_1$. At the same time,another projectile $B$ is thrown with velocity $v_2$ upwards from point $Q$,which is vertically below the highest point that $A$ would reach. For $B$ to collide with $A$,the ratio $\frac{v_2}{v_1}$ should be:

$A$ projectile is given an initial velocity of $\hat{i}+2 \hat{j} \,ms^{-1}$. The Cartesian equation of its path is ($x$ and $y$ are in metres and $g=10 \,ms^{-2}$)

$A$ player can throw a ball to a maximum horizontal distance of $80 \,m$. If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is (in $\,m$)

$A$ ball is thrown upwards at an angle of $60^o$ to the horizontal. It falls on the ground at a distance of $90 \,m$. If the ball is thrown with the same initial velocity at an angle $30^o$,it will fall on the ground at a distance of ........ $m$.

The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal is given as $y = x - \frac{gx^2}{2}$. The initial velocity of the ball is $[\text{Given} : \tan 45^{\circ} = 1, \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$