SOLUTION: The path of a falling object is given by the function s=-16t^2+v0t+s0 where v0 represents the initial velocity in ft/sec and s0 represents the initial height in feet.
If a rock i
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Question 120078: The path of a falling object is given by the function s=-16t^2+v0t+s0 where v0 represents the initial velocity in ft/sec and s0 represents the initial height in feet.
If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot building, write the height (s) equation using this information.
How high is the rock after 0.5 second?
After how many seconds will the graph reach maximum height?
What is the maximum height?
Answer by stanbon(75887) (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 v0 represents the initial velocity in ft/sec and s0 represents the initial height in feet.
If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot building, write the height (s) equation using this information.
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EQUATION:
h(t)=-16t^2+32t+40
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How high is the rock after 0.5 second?
Find h(0.5) = 52 ft
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After how many seconds will the graph reach maximum height?
Maximum height when t = -b/2a = -32/(-2*16)= 1 second
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What is the maximum height?
Find h(1) = -16+32+40 = 56 ft
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Cheers,
Stan H.
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