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Referring to the previous illustration, if the boy releases the ball from rest, what will be the radius of curvature of the path at the instant of its release? |
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Answer» Solution :The radius of curvature, `r=(V^(2))/(gostheta_(0))` `theta_(0)`=angle between v & HORIZONTAL = 0, when the ball is released. It STARTS moving horizontally with a SPEED y = speed of the train `10 m//sec`. `rArr r=(v^(2))/(gcostheta^(@))=((10)^(2))/(g)=10m` `thetA_(0)`=angle between V & horizontal = 0 |
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