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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? |
Answer» `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` |
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