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

[groetschel]

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?

 

[erobz]

Can you write how far A falls given a distance B climbs when they meet?

 

[groetschel]

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.

 

[bob012345]

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.

 

[erobz]

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.

 

[groetschel]

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

 

[bob012345]

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?

 

[groetschel]

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.

 

[bob012345]

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.

 

[groetschel]

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.

 

[erobz]

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.

 

[groetschel]

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?

 

[erobz]

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.

 

[erobz]

What is preventing you from solving for $t$?

 

[groetschel]

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?

 

[erobz]

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.

 

[groetschel]

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...

 

[erobz]

???
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?

 

[groetschel]

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.

 

[erobz]

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.

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

 

[groetschel]

Ok, let me try this...
$h - vt - \frac 1 2 g t^2 = 0$
$$ ax^2 + bx + c = 0 $$

 

[erobz]

Ok, let me try this...
$$H - vt - \frac 1 2 = 0

$$ Centered Equation $$

You forgot the set of dollar signs at the end.

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

Also, put braces in the fraction function, you will need it later when numerator and denominator are more complex.

 

[erobz]

Good. Now, in terms of variables (no values yet) what are:

$a = ?$
$b = ?$
$c = ?$

 

[groetschel]

$$ Centered Equation $$

You forgot the set of dollar signs at the end.

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

Also, put braces in the fraction function, you will need it later when numerator and denominator are more complex.

 

[erobz]

View attachment 332765

You can't do what you are doing. The righthand side is non-zero

 

[groetschel]

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?

 

[erobz]

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:

$$ ax^2 + bx+c = 0 $$

 

[groetschel]

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???

 

[berkeman]

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.

 

[erobz]

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?

 

[bob012345]

Ok, let me try this...
$h - vt - \frac 1 2 g t^2 = 0$
$$ ax^2 + bx + c = 0 $$

What's your final answer for t?

 

[PeroK]

???
Please try it on your own:
H = 2m
v = 5.833 m/s
g = 9.8 m/s²

Solve the equation for t...

You could always have checked this on a spreadsheet. You want to find $t$ such that:
$$5.833t + \frac 1 2 9.8 t^2 = 2$$You could start at $t = 0.01$ and go up in increments of $0.01$ until you get to $2$:

u	g
5.833​	9.8​
t	ut	1/2 gt^2	d
0.01​	0.05833​	0.00049​	0.05882​
0.02​	0.11666​	0.00196​	0.11862​
0.03​	0.17499​	0.00441​	0.1794​
0.04​	0.23332​	0.00784​	0.24116​
0.05​	0.29165​	0.01225​	0.3039​

 

[Lnewqban]

@groetschel
For a different approach to the same problem, please see:
https://www.physicsclassroom.com/class/1DKin/Lesson-5/Representing-Free-Fall-by-Graphs#:~:text=Representing Free Fall by Velocity-Time Graphs&text=As learned earlier, a diagonal,time graph would be diagonal.

 

[bob012345]

Ok, let me try this...
$h - vt - \frac 1 2 g t^2 = 0$
$$ ax^2 + bx + c = 0 $$

Put the first into the form of the second and identify what is a,b and c. Then use the formula given earlier for the quadratic in terms of a,b,c to solve for t (or x if that's easier).