SOLUTION: From the top of a building, a rock is thrown straight up with an intial velocity of 32ft. per second. The equation s = -16t^2 + 32t + 48 gives is the height (s) of the rocks (t) se
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-> SOLUTION: From the top of a building, a rock is thrown straight up with an intial velocity of 32ft. per second. The equation s = -16t^2 + 32t + 48 gives is the height (s) of the rocks (t) se
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Question 878366: From the top of a building, a rock is thrown straight up with an intial velocity of 32ft. per second. The equation s = -16t^2 + 32t + 48 gives is the height (s) of the rocks (t) seconds after its thrown. Find the maximum height reached by the rock. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! From the top of a building, a rock is thrown straight up with an initial velocity of 32ft. per second. The equation s = -16t^2 + 32t + 48 gives is the height (s) of the rocks (t) seconds after its thrown. Find the maximum height reached by the rock.
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s = -16t^2 + 32t + 48
complete the square:
s = -16(t^2-2t+1)+16 + 48
s = -16(t-1)^2+64
This is an equation of a parabola that opens downward with vertex at (1,80)
maximum height reached by the rock=64 ft
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