Question 95647: I hope I put the correct type of problem in my title. I am taking an online class in College Algebra and I can't figure what they are asking for in the assignments. It has been 30 years since I have done algebra. Please help me with this problem. Thank you in advance. I did solve some of the questions, I think they are right. I didn't know how to get c & d.
The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents the initial height in feet.
a) 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
Answer: s=-16t^2+64t+25
b) How high is the rock after 1 second?
Answer:s=73
Show your work here:
S=-16(1^2)+64(1)+25
S=-16+64+25
S=73
c) After how many seconds will the graph reach maximum height?
Answer:
Show your work here:
d) What is the maximum height?
Answer:
Show your work here:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a) 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
Answer: s(t)=-16t^2+64t+25
Correct
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b) How high is the rock after 1 second?
Answer:s=73
Show your work here:
S(1)=-16(1^2)+64(1)+25
S(1)=-16+64+25
S(1)=73 ft.
Correct
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c) After how many seconds will the graph reach maximum height?
Answer:
Your height function is a quadratic with a=-16, b= 64 , c=25
The maximum height occurs when t=-b/2a = -64/(2*-16) = 2 seconds
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d) What is the maximum height?
Answer:
Show your work here:
Find S(2) which is the height after 2 seconds
S(2) = -16(2^2)+64(2)+25
S(2) = -64+128+25 = 89 ft
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Cheers,
Stan H.
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