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