A stone is thrown horizontally from a height with a velocity v_(x) = 15m//s. Determine the normal and tangential acceleration of the stone in 1 second after it begins to move.

A stone is thrown horizontally from a height with a velocity v_(x) = 1

1.	A stone is thrown horizontally from a height with a velocity v_(x) = 15m//s. Determine the normal and tangential acceleration of the stone in 1 second after it begins to move.
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 g. Thus `a = g = SQRT(a_(t)^(2) + a_(N)^(2))` from figure we can 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_(1)g(v_(y))/(v) = (g^(2)t)/(sqrt(v_(x)^(2) + g^(2)t^(2))` and `a_(c) = g(v_(x))/(v) = (g v_(x))/(sqrt(v_(x)^(2) + g^(2)t^(2)))` on substituting numerical values, `v_(x) = 15m//s, g = 9.8 m//s^(2)` we get `a_(t) = 5.4 m//s^(2)` and `a_(n) = 8.2 m//s^(2)`.