The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is

$A$ particle moves in the $xy$ plane with a constant acceleration $g$ in the negative $y$-direction. Its equation of motion is $y = ax - bx^2$,where $a$ and $b$ are constants. Which of the following are correct?

$A$ particle is projected from a horizontal plane such that its velocity vector at time $t$ is given by $\vec{v} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by

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 $[\text{Given} : \tan 45^{\circ} = 1, \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

$A$ player kicks a ball at a speed of $20 \ ms^{-1}$ so that its horizontal range is maximum. Another player $24 \ m$ away in the direction of the kick,starts running in the same direction at the same instant the ball is kicked. If he has to catch the ball just before it reaches the ground,he should run with a velocity equal to: (Take $g = 10 \ ms^{-2}$)

Galileo,in his book $Two \text{ } new \text{ } sciences$,stated that "for elevations which exceed or fall short of $45^{\circ}$ by equal amounts,the ranges are equal". Prove this statement.