The equation of a projectile is $y = ax - bx^2$. Its horizontal range is ......

$A$ ball is projected from the ground with a speed $15 \, m/s$ at an angle $\theta$ with the horizontal such that its range and maximum height are equal. Then,$\tan \theta$ will be equal to:

$A$ 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:

$A$ stone is thrown vertically at a speed of $30 \,ms^{-1}$ making an angle of $45^{\circ}$ with the horizontal. What is the maximum height reached by the stone (in $\,m$)? Take $g=10 \,ms^{-2}$.

$A$ body is projected from the ground with a velocity $v = (3 \hat{i} + 10 \hat{j}) \text{ m/s}$. The maximum height attained and the range of the body respectively are (given $g = 10 \text{ m/s}^2$):

$A$ projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) \text{ ms}^{-1}$. The equation of its path is $(g=10 \text{ ms}^{-2})$