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A stone is throwns horizontally from a height with a velocity v_(x)=15 m//s. Determine the normal and tangential acceleration of the stone in I second after it begins to move, |
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Answer» Solution :The horizontal component of acceleration is ZERO. The net acceleration of the stone is DIRECTED VERTICALLY downward and is equal to the acceleration due to gravity 8. Thus`a=g= sqrt(a_(1)^(2)+a_(n)^(2))` from figure we an see that `cos theta=(v_(x))/(v)=(a_(c))/(a)=(a_(c))/(g)` and `sin theta=(v_(y))/(v)=(a_(t))/(a)=(a_(t))/(g)` Hence `a_(t)=g(v_(y))/(v)=(g^(2)t)/(sqrt(v_(x)^(2)+g^(2)t^(2)))` and `a_(c)=g(v_(x))/(v)= (gv_(x))/( sqrt(v_(x)^(2)+g^(2)t^(2)))` On SUBSTITUTING numerical values `v_(x)=15m//s anda_(a)=9.8 m//s^(2)` we get `a_(t)=5.4 m//s^(2) and a_(n)=8.2m//s^(2)` |
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