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A particle is projected wityh velocity 20 `ms^(-1)` at angle `60^(@)` above horizontal up on an inclined plane having inclination `30^(@)`. Find its shortest distance from the inclined plane when it is moving `53^(@)` above horizontal. (Using `=10ms^(-2)`)A. 11.1mB. 8.2mC. 3.3mD. 14.4m |
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Answer» `V_("vertical") U_("vertical")-"gt"` `10 tan 53^(@)=20 sin 60^(@)-10t` `(4)/(3)=sqrt(3)-1` `t=sqrt(3)-(4)/(3)` `s_(y)=u_(y)+a_(y)t^(2)` `=10xx((3sqrt(3)-4)/(3))-(1)/(2)g cos 30^(@) ((3sqrt(3)-4)/(3))^(2)` `=10xx((3sqrt(3)-4)/(3))-(1)/(2)xx10sqrt(3)/(2)((3sqrt(3)-4)/(3))^(2)` `=3.99-2.5xx1.732((3xx1.732-4)/(7))^(2)` `=3.99-4.33xx0.16` `=3.99-0.688` `=3.99-0.69=3.30` |
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