A stone is thrown horizontally with the velocity 15m/s. Determine the tangential and normal accelerations 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)`