A projectile has the maximum range of `500m`. 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?

Explanation

`R_(max)=u^(2)/g`
`:.500=u^(2)/g or u=sqrt(500g)`
`v^(2)-u^(2)=2as`
`0-500g=2xx(-gsin30^(@))x`
`x=500m`