If the equation of motion of a projectile is $y=Ax-Bx^2$,then the ratio of the maximum height reached and the range of the projectile is

$A$ projectile is given an initial velocity of $(3 \hat{i} + 4 \hat{j}) \text{ m s}^{-1}$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. Assuming $g = 10 \text{ m s}^{-2}$, if the equation of its trajectory can be written as $\frac{1}{9} [\beta x + \gamma x^2]$, then the value of $\gamma$ is

What is the range of a projectile? Also,provide the velocity of the projectile at its maximum height.

$A$ projectile is fired with a speed $u$ at an angle $\theta$ with the horizontal. Its speed when its direction of motion makes an angle $\alpha$ with the horizontal is

$A$ bullet is fired from a cannon with a velocity of $500 \, m/s$. If the angle of projection is $15^\circ$ and $g = 10 \, m/s^2$,then the horizontal range is:

$A$ ball is projected from the ground with a speed of $20\,m/s$ at an angle of $45^{\circ}$ with the horizontal. There is a wall of $25\,m$ height at a distance of $10\,m$ from the projection point. The ball will hit the wall at a height of $.........\,m$.