If the initial velocity in the horizontal direction of a projectile is the unit vector $\hat{i}$ and the equation of the trajectory is $y = 5x(1 - x)$,find the $y$-component vector of the initial velocity. (Take $g = 10\,m/s^2$) (in $,\hat{j}$)

If a body projected with a velocity of $19.6 \ m/s$ reaches a maximum height of $9.8 \ m$,then the range of the projectile is (Neglect air resistance). (in $m$)

Assertion $(A)$: The range of a projectile is maximum when the angle of projection is $45^{\circ}$.
Reason $(R)$: The range of a projectile depends only on the angle of projection.

If a body is projected with an angle $\theta$ to the horizontal,then

$A$ cricket ball is hit at $30^{\circ}$ with the horizontal with kinetic energy $K$. The kinetic energy at the highest point is

$A$ ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \ m$ high and $0.1 \ m$ wide. The minimum velocity $u$ with which the ball just hits the $5^{\text{th}}$ step of the stairway will be $\sqrt{x} \ ms^{-1}$ where $x=$ . . . . . . [use $g=10 \ ms^{-2}$].