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