Ratio between maximum range and square of time of flight in projectile motion is

The trajectory of a projectile near the surface of the earth is given as$ y = 2x -9x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$ then $(g = 10\, ms^{-2}$)

A cricket fielder can throw the cricket ball with a speed $v_{0} .$ If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal, find

$(a)$ the effective angle to the horizontal at which the ball is projected in air as seen by a spectator

$(b)$ what will be time of flight?

$(c)$ what is the distance (horizontal range) from the point of projection at which the ball will land ?

$(d)$ find $\theta$ at which he should throw the ball that would maximise the horizontal range as  found in $(iii)$.

$(e)$ how does $\theta $ for maximum range change if $u > u_0$. $u =u_0$ $u < v_0$ ?

$(f)$ how does $\theta $ in $(v)$ compare with that for $u=0$ $($ i.e., $45^{o})$ ?

Two projectiles $A$ and $B$ are thrown with the same speed such that $A$ makes angle $\theta$ with the horizontal and $B$ makes angle $\theta$ with the vertical, then

The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50\,m$. If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be $........\,m$

What is range of the projectile particle ? Give velocity of projectile particle at maximum height.