Saved Bookmarks
| 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_(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)` |
|