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 ?

Explanation

`R_("max")=(u^(2))/(g) " " :. 500 =(u^(2))/(g) or u= sqrt(500g)`
`v^(2)-u^(2)= 2gs rArr 0-500g=2xx(-g 30^(@)) xx x`
`x=500 m`