Finding when a object hits the ground, its velocity when it hits, etc

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    Ground Velocity
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Discussion Overview

The discussion revolves around determining when an object hits the ground and its velocity at that moment, based on the height function $$H(t) = 152t - 16t^2$$. The scope includes mathematical reasoning and conceptual clarification regarding the interpretation of the results.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant calculates the time when the object hits the ground as $$t = \frac{152}{16}$$ and notes that this implies two solutions: $$t = 0$$ and $$t = \frac{152}{16}$$.
  • Another participant confirms the calculation and notes that the final and initial velocities should have the same magnitude but opposite directions, emphasizing the negative sign indicates downward motion.
  • A different participant questions why the time to hit the ground was not reduced to $$\frac{19}{2}$$ or 9.5 seconds, suggesting clarity in presenting the answer.
  • One participant points out that while $$t = 0$$ is a solution, it is not relevant in this context since it represents the initial position of the object, which is at ground level.
  • Another participant agrees with the importance of recognizing which roots of the equation are applicable in the context of the problem.

Areas of Agreement / Disagreement

Participants generally agree on the calculations for when the object hits the ground and the velocity at that moment, but there is a disagreement regarding the relevance of the time $$t = 0$$ in the context of the problem.

Contextual Notes

There is an emphasis on the importance of interpreting the roots of the equation correctly, particularly in distinguishing between the initial position and the time when the object actually hits the ground.

shamieh
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I need someone to check my work, because I am getting very weird numbers. Just want to make sure I am doing this right...
The height of an object in meters after t seconds is given by $$H(t) = 152t - 16t^2$$.

a) When does it hit the ground?
My answer: t(152 - 16t) = 0 which means t = 0, $$t = \frac{152}{16}$$

b) What is the velocity when it hits?

152 - 304 = -152 m/s

Is this correct?
 
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Yes, that's correct...you should expect in this case that the final and initial velocities have the same magnitude, but opposite directions. The negative sign on the final velocity means it is traveling in a downward direction.
 
shamieh said:
I need someone to check my work, because I am getting very weird numbers. Just want to make sure I am doing this right...
The height of an object in meters after t seconds is given by $$H(t) = 152t - 16t^2$$.

a) When does it hit the ground?
My answer: t(152 - 16t) = 0 which means t = 0, $$t = \frac{152}{16}$$
Any particular reason why you didn't reduce this to \dfrac{19}{2} or 9.5? Also it should be 19/2 seconds or 9.5 seconds.

b) What is the velocity when it hits?

152 - 304 = -152 m/s

Is this correct?
H'= 152- 32t
152- 32(19/2)= 152- 304= -152 m/s
 
shamieh said:
a) When does it hit the ground?
My answer: t(152 - 16t) = 0 which means t = 0, $$t = \frac{152}{16}$$
A picky, but important point. t = 0 s is not an answer here. It is at the ground at t = 0 s, but is being fired upward at that point.

-Dan
 
topsquark said:
A picky, but important point. t = 0 s is not an answer here. It is at the ground at t = 0 s, but is being fired upward at that point.

-Dan

Good catch, Dan! :D

I don't think that's picky at all. Students should learn early on which roots of their equations apply and which do not and why. (Yes)
 

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