Q: A projectile has the maximum range of 500 m. if the projectile is now thrown up on an inclined plane of 30° with the same speed, what is the distance covered by it along the inclined plane ?
Sol: $\large R_{max} = \frac{u^2}{g}$
$\large 500 = \frac{u^2}{g}$
$\large u^2 = 500 g $
$\large v^2 = u^2 + 2as $
$\large 0 = 500 g + 2(-gsin30^o)s $
s = 500 m