The equation of motion of a projectile is $y = Ax -Bx^2$ where $A$ and $B$ are the constants of motion. The horizontal range of the projectile is
$\frac{A}{B}$
$\frac{B}{A}$
$\frac{A^2}{B}$
$\frac{B^2}{A}$
Define projectile particle and derive the equation $y\, = \,(\tan \,{\theta _0})x\, - \,\frac{g}{{(2\,\cos \,{\theta _0})}}{x^2}$
The maximum horizontal range of a projectile is $400\, m$. The maximum value of height attained by it will be ......... $m$
If we can throw a ball upto a maximum height $H$, the maximum horizontal distance to which we can throw it is
A ball is projected with velocity ${V_o}$ at an angle of elevation $30^°$. Mark the correct statement
A boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\,m/s$ at an angle of $30^o$ with the horizontal. How far from the throwing point will the ball be at the height of $10\, m$ from the ground ? $\left[ {g = 10\,m/{s^2},\sin \,{{30}^o} = \frac{1}{2},\cos \,{{30}^o} = \frac{{\sqrt 3 }}{2}} \right]$