Thrown Ball on Planet Zeno: Find Height & Time to Ground

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Homework Help Overview

The problem involves a ball thrown vertically upward on planet Zeno, described by the height function h(t) = 45t - 9t². Participants are tasked with determining the time it takes for the ball to hit the ground and the maximum height it reaches.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the implications of the height function, questioning the initial velocity and acceleration values. Some suggest using derivatives or graphing to analyze the function, while others discuss the vertex of the parabola. There are also considerations about the units used in the equations.

Discussion Status

The discussion includes various interpretations of the height function and its parameters. Some participants provide guidance on finding roots and maximum height, while others express confusion about the units and the method of completing the square. There is no explicit consensus on the approach, but several productive lines of reasoning are being explored.

Contextual Notes

Participants note potential confusion regarding the initial conditions and units of measurement, which may affect the interpretation of the problem. There is also a mention of homework constraints that may limit the methods discussed.

mustang
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Problem 9. If a ball is thrown vertically upward on the planet Zeno at 100m/s, then it's approximate height in meters t seconds later is given by h(t)=45t-9t^2. a.After how many seconds does the ball hit the ground?
b. How high does the ball go?
 
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What is the next value of t such that h(t) is equal to zero?

Have you studied derivatives? If so, where is the derivative of h(t) equal to zero?

If not, h(t) describes a parabola. What is the vertex of it?

And if all else fails, try graphing it.

cookiemonster
 
h(t) = 45t - 9t2
Wouldn't that imply an initial velocity of 45m/s (and an acceleration of 9m/s2 downward), rather than 100m/s? :confused:

(You can ignore this if you want, mustang, as I don't wish to confuse you and this has nothing to do with the anwer to your problem. :smile:)
 
Assuming planet Zeno is not in some alternate universe where our traditional laws of physics do not apply, then h(t) = y = h(t) = 45t - 9t2 should indicate an initial velocity of 45 m/s and an acceleration of -18 m/s2 (where negative is downward).

[tex]y = v_0 t + \frac{1}{2}at^2[/tex]

In the given, the value -9 is equivalent to the expression [tex]\frac{1}{2} a[/tex]

[tex]-9 = \frac{1}{2}a[/tex]

Solving for a yields -18 m/s2


Sorry for the edits. I'm still learning the tex tags.
 
Last edited:
I suppose that in the equation:
h(t) = 45t - 9t2
t = 0 is not the moment in which the ball was thrown. So that does away with my stupid question. :wink:
 
It still implies an initial velocity of 45 in some units. We don't know that those units are m/s, though...

cookiemonster
 
cookiemonster said:
It still implies an initial velocity of 45 in some units. We don't know that those units are m/s, though...
We do, since it says "height in meters t seconds later is given by". If the inital velocity was 45mph the equation would be (neglecting acceleration for this purpose):
h [meters] = 45 [miles]/[hour] * t [seconds]
That is nonsensical. :-p
 
These units are killing me these days. =\ I totally skipped the specification of units for t and h.

cookiemonster
 
You look pale, have some more cookies. :smile:
 
  • #10
That's true, I haven't had any caffeine or chocolate in a while... Must be withdrawal.

cookiemonster
 
  • #11
Cookies sound good. Got milk?
 
  • #12
So if the acceleration is -18m/s does that mean that the number of seconds does the ball hit the ground is by using synthetic division and then from that use that number for t to put it in the equation to get the height?
 
  • #13
What does synthetic division have to do with this? We're not dividing polynomials here.

All you have to do is factor and find the roots of the equation. This is algebra 1 stuff.

cookiemonster
 
  • #14
h(t)=45t-9t^2
0=9t(5-t)
t=0 or t=5
So the time is five seconds.

So then would the value of t=5 in t.
The height would be 0?
 
  • #15
Well, naturally. That's when the function is zero and when the ball hits the ground.

If you want to find the highest point, you have to complete the square. More algebra 1.

cookiemonster
 
  • #16
The height

SO by completing the square:
It would be -9t^2+45t=0
(.5*45)^2=2054/4
-9t^2+45t+2054/4=2054/4
Is this right??
 
  • #17
Remember that in order to complete the square, the coefficient of the t^2 term must be 1.

You can actually use your intuition and figure this out. It's a parabola, so it's going to be symmetric. It has a value of 0 at 0 and 5s. Sometime between those two points of time, it attains its maximum. At what time do you suppose it'll be a max?

The parabola's symmetric... Symmetric...

cookiemonster
 
  • #18
Regarding the height

Since it stays in the air for 5 seconds midway at 2.5 seconds it reaches it highest height of -9(2.5)^2+45(2.5)=56.25. Right??
 
  • #19
Correct.

cookiemonster
 
  • #20
A Big Thank You!

Thanks to all of you guys especially "COOKIEMONSTER."
 

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