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

Explanation

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
`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) =gv_(y)/v =(g^(2)t)/sqrt(v_(x)^(2) +g^(2)t^(2))`
and `a_( c) gv_(s)/v =(gv_(x))/sqrt(v_(x)^(2) + g^(2)t^(2))`
on substituting numerical values,
`v_(x) =15 m//s, g=9.8 m//s^(2)`
we get, `a_( t)=5.4 m//s^(2)` and `a_(n) =8.2 m//s^(2)`