The maximum range of a projectile is 500m. If the particle is thrown up a plane is inclined at an angle of `30^(@)` with the same speed, the distance (in meter) covered by it along the inclined plane will be:

Answer

500

Explanation

For the maximum range, `theta=45^(@)`
`R=(u^(2)sin 2 theta)/g=(u^(2))/g sin 90^(@)=(u^(2))/g`
or `500=(u^(2))/g`
The distance covered along the inclined plane can be obtain using the equation
`v^(2)-u^(2)=2as`
or `0-u^(2)=2(-g sin 30^(@))s`
or `s=(u^(2))/g=500m`