For how long will the ball rise?

D

draginto

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When a ball is thrown straight up from the top of a building into the air, its position may be measured as the vertical distance from the ground. Regard "up" as the positive direction and let s(t) be the height of the ball in feet after t seconds. The movement of the ball is given by the function s(t)=128t-16t^2+144.

e. For how long will the ball rise?
f. How far did it rise?
 
Hello, and welcome to FMH! :)

When the ball reaches the apex of its trajectory, what do we know about its instantaneous velocity?
 
Thank you for your greeting and answer!
Hmmm, I would say, that the instantaneous velocity is the derivative of the position function?
 
draginto said:
Thank you for your greeting and answer!
Hmmm, I would say, that the instantaneous velocity is the derivative of the position function?
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Yes, that's correct, and will help us, but first think about the ball as it rises and then falls...what must be the magnitude of its velocity at that moment when it is at the apex, as it is neither rising nor falling...
 
Correct, and so use \(s(4)-s(0)\) to answer the second part of the posted question...
 
The velocity function of the ball is continuous, that is it can't skip any speed during its flight. Now when the ball is going up we can call this positive velocity and when the ball is falling it is negative velocity. Now to go from a positive velocity to a negative velocity the ball had to be zero somewhere. This velocity is zero at the max height.
At some point you will learn about the intermediate value theorem which talks about what I said above.
 
draginto said:
So the answer to the second part of the question would be s=400ft?
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No, that would be its height, but recall it began 144 ft. up, in other words you have to subtract its initial height to get the difference in height, or how far it rose.
 
That is why MarkFL said s(4) - s(0). Note that s(0) is the initial height of the ball.
 
Thank you all, I learned much from the process. I really appreciate the time and patience given to me towards answering this question.
 
draginto said:
Yes thank you, I had missed the s(0). Which is where I made the mistake.
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I initially only posted \(s(4)\) but edited my post within seconds to write the difference \(s(4)-s(0)\), but you may have seen it before I edited. The change in height is the final height minus the initial height. If you take a course in physics, you'll learn an easier way to find the change in height, based on energy, but it's good to be able to solve problems via dynamics like this.
 
MarkFL said:
I initially only posted \(s(4)\) but edited my post within seconds to write the difference \(s(4)-s(0)\), but you may have seen it before I edited. The change in height is the final height minus the initial height. If you take a course in physics, you'll learn an easier way to find the change in height, based on energy, but it's good to be able to solve problems via dynamics like this.
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Oh, this may have been the case. Still, I'm glad I was able to figure it out eventually through your help. I may take a physics course, or self-study. It's definitely on the agenda. The application of derivatives kind of problems however, always seem to confuse me. Is there something you would recommend for getting a better grasp at understanding them?
 
draginto said:
...The application of derivatives kind of problems however, always seem to confuse me. Is there something you would recommend for getting a better grasp at understanding them?
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At the risk of being cliched, practice practice practice. Differential calculus is especially useful in the natural sciences, in particular physics. When I was a student, problems that related to physics, that is to situations that we can easily intuit where things change with time or position, made the derivative come alive for me. :)
 
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