New, and in desperate need of help.

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desperatha

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Nov 30, 2011
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This is the problem: A ball is thrown into the air from the top of a 20 m tall building at an initial velocity of 10 m/s. The ground below is taken to be 0 m. The height of the ball can be be approximated by the following function:

h(t) = h0 + v0t - 10t^2

-Write the function algebraically.
-What is the maximum height and the time it takes to reach that height (vertex)?
-What is the y-axis intercept of the function?
-Find the zeroes of the function.
-What is the height after 5 seconds have elapsed?
-What is the function algebraically?
-What is the y-axis intercept of the function?

I'm at a complete loss as to how to even attempt solving this. I think, algebraically, the function would be h(t) = 20 + 10t - 10t^2 - assuming I plugged everything in correctly? :(
 
desperatha said:
This is the problem: A ball is thrown into the air from the top of a 20 m tall building at an initial velocity of 10 m/s. The ground below is taken to be 0 m. The height of the ball can be be approximated by the following function:

h(t) = h0 + v0t - 10t^2

-Write the function algebraically. Yes, you are correct- h(t) = 20 + 10t - 10t^2
-What is the maximum height and the time it takes to reach that height (vertex)? The max occurs at h'(t) = 0. So differentiate and find solutions to h'(t) = 0. If you do not know how to differentiate h'(t) =10-20t. Now you know the time, so take that and plug it into the original question h(t). You now have your time and height.
-What is the y-axis intercept of the function? This occurs when t=0 or h(0).
-Find the zeroes of the function. This occurs when h(t) = 0, you will get two answer. NOTE: since it asks for the zeroes, i guess you should include both the neg and positive answer. However, negative time is usually discarded.
-What is the height after 5 seconds have elapsed? h(5) - don't forget where the ground is though.
-What is the function algebraically? Quadratic or degree 2 polynomial.
-What is the y-axis intercept of the function? Initial height = the height of the building.

I'm at a complete loss as to how to even attempt solving this. I think, algebraically, the function would be h(t) = 20 + 10t - 10t^2 - assuming I plugged everything in correctly? :(
Click to expand...
^^
 
Last edited:
Thanks for taking the time to help me :eek:

If I'm understanding this right, I get -10 for the time (10 - 20), and the equation for the height would be 20 + 10(-10) - 10(-10)^2?
 
desperatha said:
For finding the maximum height, and the time it takes to reach that height.
Click to expand...

-What is the maximum height and the time it takes to reach that height (vertex)? The max occurs at h'(t) = 0. So differentiate and find solutions to h'(t) = 0. If you do not know how to differentiate h'(t) =10-20t. Now you know the time, so take that and plug it into the original question h(t). You now have your time and height.
Click to expand...
Solve 0=10-20t

That will be your time for max height.

Then plug that time into the original equation to get the height.
 
desperatha said:
h = 22.5 m?

and

h(5) = - 180?
Click to expand...

Yes, the maximum height is 22.5 m, and that's reached after 1/2 second.

h(0.5) = 22.5

h(5) = -180 would mean that the object is traveling through the ground on its way toward the center of the earth and is located 180 m below the surface 5 seconds after it was tossed into the air.

Do you know? The formula for the x-coordinate of the parabola's vertex is x = -b/(2a). Of course, it is t = -b/(2a) in this exercise.
 
mmm4444bot said:
Yes, the maximum height is 22.5 m, and that's reached after 1/2 second.

h(0.5) = 22.5

h(5) = -180 would mean that the object is traveling through the ground on its way toward the center of the earth and is located 180 m below the surface 5 seconds after it was tossed into the air.

Do you know? The formula for the x-coordinate of the parabola's vertex is x = -b/(2a). Of course, it is t = -b/(2a) in this exercise.
Click to expand...

:? I was told, by my Algebra teacher, that the answer to that problem was going to be a negative number (even though that implies the ball is going through the ground), because she didn't fully think out the problem when she came up with it.
 
desperatha said:
:? I was told, by my Algebra teacher, that the answer to that problem was going to be a negative number (even though that implies the ball is going through the ground), because she didn't fully think out the problem when she came up with it.
Click to expand...

Classic! I love these teachers that do not work out the problems themselves before giving them to their students. Even better is when the teachers give the students a key to some sample quiz or study material and there are incorrect answers on the key. I see it way too many times as a tutor. The kids that are on the fence on how to do some of these problems may very well get the correct answer but their answer does not match the answer key since the key's answer is wrong so then the kid assumes, of course, that he is wrong. Sad.
 
desperatha said:
:? I was told, by my Algebra teacher, that the answer to that problem was going to be a negative number (even though that implies the ball is going through the ground), because she didn't fully think out the problem when she came up with it.
Click to expand...

The problem does state that the ground is at 0 m.

I just assumed the teacher was testing ones ability to read the question properly.

My best interpretation would be at anything after t=2 (x-int) the ball is at height 0 m.

The question is somewhat ambiguous though. I mean, does she want us to ignore the ground? does she also want the negative time for the zeros? - these are nonsensical answers for the problem.

She should be testing the ability to discard extraneous roots and about practical domain.
 
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