Object A (in free fall) hits object B (ascending with constant velocity)

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SUMMARY

The discussion centers on solving a quadratic equation derived from the motion of two objects: Object A, which falls from a height of 2 meters with an acceleration of 9.8 m/s², and Object B, which ascends at a constant velocity of 5.833 m/s. The key equation established is 2m - 1/2*9.8m/s²*t² = 5.833m/s*t. Participants emphasize the importance of defining a coordinate system and using the quadratic formula to find the time of impact. The final result for time, t, is calculated to be approximately 1.19 seconds, but discrepancies in distance traveled raise concerns about the accuracy of this solution.

PREREQUISITES
  • Understanding of kinematic equations, specifically for free fall and constant velocity.
  • Familiarity with quadratic equations and the quadratic formula.
  • Basic knowledge of coordinate systems and how to define positive and negative directions in physics.
  • Ability to use LaTeX for formatting mathematical expressions.
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  • Learn how to apply the quadratic formula to solve for time in motion problems.
  • Study the concept of coordinate systems in physics to better visualize motion scenarios.
  • Explore methods for graphically representing free fall and constant velocity, including velocity-time graphs.
  • Practice using LaTeX for mathematical expressions to improve clarity in problem-solving discussions.
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Students and educators in physics, particularly those studying kinematics, as well as anyone involved in solving motion-related problems in a mathematical context.

groetschel
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Homework Statement
Hey, I am lost.
A falls from a height of 2m with an acceleration of 9.8m/s2 down (free fall).
B ascends with a constant speed of 5.833 m/s
When do A and B hit each other?
Thank you very much for the help.
Gunnar
Relevant Equations
DofA = D0 + 1/2at^2
DofB = vt
DofA = DofB, but how do I solve this?
DofA = D0 + 1/2at^2
DofB = vt
DofA = DofB, but how do I solve this and get the time when A and B hit?
 
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Can you write how far A falls given a distance B climbs when they meet?
 
erobz said:
Can you write how far A falls given a distance B climbs when they meet?
No, I am sorry. What I have given is all I have. In the end, it is just a quadratic function I have to solve for t, but my result is only crap.
 
First define a coordinate system. Where is zero? Where are A and B in that system. Draw a little sketch. Then you will see clearly how to set up the problem.
 
groetschel said:
No, I am sorry. What I have given is all I have.
Draw a diagram. That shows B climbing, A falling. You should notice a relationship that can be made when the meet.
 
bob012345 said:
First define a coordinate system. Where is zero? Where are A and B in that system. Draw a little sketch. Then you will see clearly how to set up the problem.
B starts at y=0 straight up
A starts at a height of 2m directly above B and falls with 9.8m/s2
 
groetschel said:
B starts at y=0 straight up
A starts at a height of 2m directly above B and falls with 9.8m/s2
Ok. You have something close in your OP but need to think through the physical picture clearly (the physics). Hint, which direction is positive and which is negative in your system? Do the equations reflect that?
 
bob012345 said:
Ok. You have something close in your OP but need to think through the physical picture clearly (the physics). Hint, which direction is positive and which is negative in your system? Do the equations reflect that?
Yes, I would say so. B climbs: positive, A falls: negative.
 
groetschel said:
Yes, I would say so. B climbs: positive, A falls: negative.
Ok, good. Write an equation for A and B with the proper directions in mind.
 
  • #10
bob012345 said:
Ok, good. Write an equation for A and B with the proper directions in mind.
A: Distance = 2m - 1/2*9.8m/s2*t2
B: Distance = 5.833m/s*t

A: Distance = B: Distance
2m - 1/2*9.8m/s2*t2=5.833m/s*t

The question is now: how do I solve this quadratic function? All my attempts do not make sense.
 
  • #11
groetschel said:
A: Distance = 2m - 1/2*9.8m/s2*t2
B: Distance = 5.833m/s2*t

A: Distance = B: Distance
2m - 1/2*9.8m/s2*t2=5.833m/s2*t

The question is now: how do I solve this quadratic function? All my attempts do not make sense.
Draw I diagram. You’ll notice something that isn’t correct. Also, you have the units of velocity incorrect.
 
  • #12
erobz said:
Draw I diagram. You’ll notice something that isn’t correct. Also, you have the units of velocity incorrect.
Yeah, you are right. the units of the velocity are incorrect - I corrected them.
I drew a diagram, but what is wrong? The acceleration works downwards to earth, the speed upwards... Where can I go wrong?
 
  • #13
groetschel said:
Yeah, you are right. the units of the velocity are incorrect - I corrected them.
I drew a diagram, but what is wrong? The acceleration works downwards to earth, the speed upwards... Where can I go wrong?
Yeah, never mind what I said.
 
  • #14
What is preventing you from solving for ##t##?
 
  • #15
erobz said:
What is preventing you from solving for ##t##?
2m - 1/2*9.8m/s2*t2=5.833m/s2*t
1/2*9.8m/s2 * t2 + 5.833m/s * t = 2m
t2 + 2*5.833m/s /9.8m/s2 * t = 4m
t (t + 1.19s)2 = 4m

Something is wrong, isn't it?
 
  • #16
groetschel said:
2m - 1/2*9.8m/s2*t2=5.833m/s2*t
1/2*9.8m/s2 * t2 + 5.833m/s * t = 2m
t2 + 2*5.833m/s /9.8m/s2 * t = 4m
t (t + 1.19s)2 = 4m

Something is wrong, isn't it?
Try using the quadratic formula, (I'm not sure what you are trying to do). Work in all variables for the moment so its not so messy. Start with:

$$ H - vt -\frac{1}{2}gt^2 = 0 $$

Please see LaTeX Guide for formatting math in posts like above.
 
  • #17
erobz said:
Try using the quadratic formula, (I'm not sure what you are trying to do). Work in all variables for the moment so its not so messy. Start with:

$$ H - vt -\frac{1}{2}gt^2 = 0 $$

Please see LaTeX Guide for formatting math in posts like above.
???
Please try it on your own:
H = 2m
v = 5.833 m/s
g = 9.8 m/s2

Solve the equation for t...
 
  • #18
groetschel said:
???
Please try it on your own:
H = 2m
v = 5.833 m/s
g = 9.8 m/s2

Solve the equation for t...
I have done it, what is the issue you are having?
 
  • #19
erobz said:
I have done it, what is the issue you are having?
My result is 1.19s

1.19s does not make sense, because B will travel D = v*t = 5.833 m/s * 1.19s = 6.9 m
This does not make sense, because the distance between A and B is only 2 m.
 
  • #20
I don't get that time, you are making an error somewhere. If you're having difficulties with the quadratic formula you could try completing the square? Or try showing me the numbers you are plugging in, but do it in Latex please; LaTeX Guide. Its not difficult to learn and it makes examining your equations easier.

## t = \frac{-b \pm \sqrt{b^2 - 4 a c }}{2a} ##
 
  • #21
Ok, let me try this...
##h - vt - \frac 1 2 g t^2 = 0##
$$ ax^2 + bx + c = 0 $$
 
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  • #22
groetschel said:
Ok, let me try this...
$$H - vt - \frac 1 2 = 0

Code: 
$$ Centered Equation $$

You forgot the set of dollar signs at the end.

Code: 
$$H - vt - \frac{1}{2}gt^2 = 0 $$

Also, put braces in the fraction function, you will need it later when numerator and denominator are more complex.
 
  • #23
Good. Now, in terms of variables (no values yet) what are:

##a = ?##
##b = ?##
##c = ? ##
 
  • #24
erobz said:
Code: 
$$ Centered Equation $$

You forgot the set of dollar signs at the end.

Code: 
$$H - vt - \frac{1}{2}gt^2 = 0 $$

Also, put braces in the fraction function, you will need it later when numerator and denominator are more complex.
IMG_0098.jpg
 
  • #25
  • #26
erobz said:
You can't do what you are doing. The lefthand side is non-zero
Then I have no idea how to solve this... Could you give me a hint?
 
  • #27
groetschel said:
Then I have no idea how to solve this... Could you give me a hint?
Have you never heard about the quadratic formula, or the method of completing the square?

If you haven't...the quadratic formula I gave to you in post #20.

$$ x = \frac{-b \pm \sqrt{b^2 - 4 a c }}{2a} $$

is a general solution to:

$$ ax^2 + bx+c = 0 $$
 
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  • #28
erobz said:
Have you never heard about the quadratic formula, or the method of completing the square?

If you haven't...the quadratic formula I gave to you in post #20.

$$ x = \frac{-b \pm \sqrt{b^2 - 4 a c }}{2a} $$

is a general solution to:

$$ ax^2 + bx+c = 0 $$
You mean something like this???
IMG_0099.jpg
 
  • #29
groetschel said:
You mean something like this???
Please post math using LaTeX. I'll send you a PM to help you get up to speed on LaTeX.
 
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  • #30
groetschel said:
You mean something like this???
View attachment 332766
It's "like that", but you've made some mistake. You've added ##\frac{v^2}{g^2}## to one side of the equation and subtracted it from the other?
 
Last edited:

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