If the instantaneous velocity of a particle projected as shown in the figure is given by $v = a \hat{i} + (b - ct) \hat{j}$,where $a, b$,and $c$ are positive constants,the range on the horizontal plane will be

The equation of motion of a projectile is $y = ax - bx^2$,where $a$ and $b$ are constants. Match the Column-$I$ with Column-$II$:

Column-$I$	Column-$II$
$i)$ The initial velocity of projection	$a)$ $\sqrt{\frac{g(1+a^2)}{2b}}$
$ii)$ The horizontal range of projectile	$b)$ $\frac{a}{b}$
$iii)$ The maximum height attained by projectile	$c)$ $\frac{a^2}{4b}$
$iv)$ The time of flight of projectile	$d)$ $a\sqrt{\frac{2}{bg}}$

$A$ projectile with speed $50 \ m \ s^{-1}$ is thrown at an angle of $60^{\circ}$ with the horizontal. The maximum height that can be reached is,(acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m$)

$A$ boy can throw a stone up to a maximum height of $10 \ m$. The maximum horizontal distance that the boy can throw the same stone up to will be .......... $m$.

$A$ projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall of height $H$ at its highest point,as shown in the figure. The horizontal distance from the point of projection to the wall is $\sqrt{3} H$. The angle $\theta$ at which the projectile is thrown is given by:

Define a projectile and derive the equation of its trajectory: $y = (\tan \theta_0)x - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$.