SOLUTION: An object moving vertically is at the given heights at the specified times. Find the position equation s =1/2at^2 + v0t + s0 for the object.
At
t = 1 second, s = 156 feet
At
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: An object moving vertically is at the given heights at the specified times. Find the position equation s =1/2at^2 + v0t + s0 for the object.
At
t = 1 second, s = 156 feet
At
Log On
Question 1178109: An object moving vertically is at the given heights at the specified times. Find the position equation s =1/2at^2 + v0t + s0 for the object.
At
t = 1 second, s = 156 feet
At
t = 2 seconds, s = 108 feet
At
t = 3 seconds, s = 28 feet Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! s=-16t^2+vt+s0
156=-16+v+s0; v+s0=172
108=-64+2v+s0; 2v+s0=172; therefore v=0 and s0=172
28=-144+3v+s0; 3v+s0=172
The equation is s(g)=-(1/2)at^2+172
It is dropped from a height of 172 feet.
