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Question 352293: An object is thrown straight up from the top of a 40 foot building at an initial rate of 72 feet per second.
a) Write the position function indicated.
b) Find the objects maximum height.
c) When does the object hit the ground? Hint: s(t)=0
d) When is it 80 feet above the ground? Hint: s(t)=80
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! a) s(t) = -16t^2 + 72t + 40 ___ gravitational acceleration(t^2) + initial velocity(t) + initial height
b) the graph of the function s(t) is a parabola , with the max height at the vertex (on the axis of symmetry)
___ the general equation for the axis of symmetry (using ax^2+bx+c) is ___ x = -b/(2a)
___ find t using the axis of symmetry , and substitute to solve for s
c) using the equation from (a) , find t when s equals 0 ___ the negative solution is extraneous
d) using the equation from (a) , find t when s equals 80
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