SOLUTION: The path of a falling object is given by the function s=-16t2+v0t+s0 where v0 represents the initial velocity in ft/sec and s0 represents the initial height in feet. If a rock is

Algebra ->  Human-and-algebraic-language -> SOLUTION: The path of a falling object is given by the function s=-16t2+v0t+s0 where v0 represents the initial velocity in ft/sec and s0 represents the initial height in feet. If a rock is      Log On


   

Question 111135: The path of a falling object is given by the function s=-16t2+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 64 feet per second from the top of a 25-foot building, write the height (s) equation using this information. Typing hint: Type t-squared as t^2
How high is the rock after 1 second?
After how many seconds will the graph reach maximum height?
What is the maximum height?
Can you show me details. thanks
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
s=-16t%5E2%2Bv%5B0%5Dt%2Bs%5B0%5D
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s=-16t%5E2%2B64t%2B25: substituting the values of v%5B0%5D=64+fps and s%5B0%5D=25feet
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After 1 second, s=-16%281%29%5E2%2B64%281%29%2B25=-16%2B64%2B25=73
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s=-16t%5E2%2B64t%2B25 would graph as a parabola opening downward, and the maximum height would occur at the vertex of the parabola. The t-coordinate of the vertex is given by -b%2F2a=-64%2F%282%28-16%29%29=2.
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s=-16%282%29%5E2%2B64%282%29%2B25=-64%2B128%2B25=89feet
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graph%28400%2C400%2C-0.5%2C6%2C-10%2C100%2C-16x%5E2%2B64x%2B25%2C25%29
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When interpreting the graph, remember that the x-axis is time, and the y-axis is distance. Notice that the graph crosses the y-axis at 25 feet because that is the value of s%5B0%5D. Also, anything on the graph to the left of the y-axis is meaningless because that would represent negative time values, and anything below the x-axis represents something beneath the ground.
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