SOLUTION: The path of a falling object is given by the function s=16t^2+v0t+s0 where Vo represents the initial velocity in ft/sec and s0 represents the initial heigth in feet.
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-> SOLUTION: The path of a falling object is given by the function s=16t^2+v0t+s0 where Vo represents the initial velocity in ft/sec and s0 represents the initial heigth in feet.
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Question 95066: The path of a falling object is given by the function s=16t^2+v0t+s0 where Vo represents the initial velocity in ft/sec and s0 represents the initial heigth in feet.
If a rock is throne upward with an initial velocity of 64 ft per sec. from the top of a 25 ft building, write the height (s) equation using this information.
a) how high is the rock after 1 sec?
b)after how many sec will the graph reach max heigth?
c) what is the max. heigth? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The path of a falling object is given by the function s=16t^2+v0t+s0 where Vo represents the initial velocity in ft/sec and s0 represents the initial height in feet.
If a rock is throne upward with an initial velocity of 64 ft per sec. from the top of a 25 ft building, write the height (s) equation using this information.
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s = -16t2 + 64t + 25
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Reviewing the Three parts of this equation
-16t^2 is the force of gravity, hence it is negative
64t is the initial upward velocity, it is positive
25 is the initial height of the building where this starts
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a) how high is the rock after 1 sec?
Substitute 1 for t in the above equation and find s:
s = -16(1^2) + 64(1) + 25
s = -16 + 64 + 25
s = 73 ft above ground after 1 second
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b)after how many sec will the graph reach max height?
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Max height will occur at the axis of symmetry of this equation:
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Formula for that is: x = -b/(2a)
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In our equation this would be:
s = -64/(2*-16)
s = -64/-32
s = +2 seconds to reach max height
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c) what is the max. height?
Find the this by substituting 2 for t in the equation:
s = -16(2^2) + 64(2) + 25
s = -16(4) + 128 + 25
s = -64 + 128 + 25
s = 89 ft is the max height occurring after 2 second
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A graphical presentation would show this clearly.
x axis in seconds and y axis in feet
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Note after 1 sec it looks like it is about 70 ft and after 2 sec it's over 88 ft
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Did I succeed in making this understandable to you?
