A stone is thrown horizontally from a height with a velocity v_9x)=15m//s. Determine the normal and tangetial acceleartion of the stone in 1 second after begins to move.

A stone is thrown horizontally from a height with a velocity v_9x)=15m

1.	A stone is thrown horizontally from a height with a velocity v_9x)=15m//s. Determine the normal and tangetial acceleartion of the stone in 1 second after 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_(1)^(2)+a_(N)^(2))` from figure we can see that `costheta = (V_("X"))/(V) = (a_( c))/(a) = (a_(g))/(g)` and `sintheta = (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, g = 9.8m//s^(2)`, we get `a_(t)=5.4m//s^(2) and a_(n) = 8.2m//s^(2)`