Question 205111: I am an online college student. The name of my book is Blitzer College Algebra 5th edition. This is a student's solution manual. The author is Prentice Hall/Pearson.
Please help with this problem. I did get part of the solution, but I am not sure how to get the last two parts. This is the problem and what I have so far.
The path of a falling object is given by the function s=-16t+vot+so where vo represents the initial velocity in ft/sec and so represents the initial height.The variable t is time in seconds, and s is the height of the object in feet.
a) If a rock is thrown upward with an initial velocity of 32 feet per second from the top os a 40-foot building, write the height equation using this information.
answer: s=-16t^2+32t+40
b) How high is the rock after 0.5 seconds?
answer: 52 feet solution: s=-16(0.5)^2+32(0.5)+40
s=-16(0.25)+32(0.5)+40
s=-4+32(0.5)=40
s=-4+16+40
s=12+40
s=52 feet
These next two questions pertain to the same problem, but I am not sure how to get the equations that need to find my answers.
c) After how many seconds will the rock reach maximum height?
d) What is the maximum height?
Your help will be greatly appreciated. Thank you, Rae
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The path of a falling object is given by the function s=-16t+vot+so where vo represents the initial velocity in ft/sec and so represents the initial height.The variable t is time in seconds, and s is the height of the object in feet.
a) If a rock is thrown upward with an initial velocity of 32 feet per second from the top os a 40-foot building, write the height equation using this information.
answer: s=-16t^2+32t+40
b) How high is the rock after 0.5 seconds?
answer: 52 feet solution: s=-16(0.5)^2+32(0.5)+40
s=-16(0.25)+32(0.5)+40
s=-4+32(0.5)=40
s=-4+16+40
s=12+40
s=52 feet
These next two questions pertain to the same problem, but I am not sure how to get the equations that need to find my answers.
c) After how many seconds will the rock reach maximum height?
If you use derivatives, set the 1st derivative to zero.
h(t) = -16t^2 + 32t + 40
h' = -32t + 32 = 0
t = 1 second.
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Without using derivatives:
Find the time when the rock is back at 40 feet.
-16t^2 + 32t + 40 = 40
t = 2 seconds.
Since rise time to max and the time back to 40' is equal, the time to max ht. is 1/2 of the 2 seconds.
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d) What is the maximum height?
max ht = h(1) = -16 + 32 + 40
= 56 feet.
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! s(t) =-16t^2+32t+40
a = -16 ; b = 32 ; c = 40
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These next two questions pertain to the same problem, but I am not sure how to get the equations that need to find my answers.
c) After how many seconds will the rock reach maximum height?
Maximum height occurs at t = -b/2a = -32/(2*-16) = 1 second
d) What is the maximum height?
The height when t = 1 is s(1) = -16+32+40 = 56 ft.
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Cheers,
Stan H.
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