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 ?

A projectile has the maximum range of 500 m. If the projectile is now

1.	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 ?
Answer» Solution :`R_("max") = (u^(2))/(g)` `THEREFORE 500 = (u^(2))/(g)` or `u = sqrt(500g)` `v^(2) - u^(2) = 2gs` `RARR 0 - 500 g = 2 XX (-g sin 30^(@)) xx x` x = 500 m.