Objects Flying towards each other (2d kinematics)

[chimpfunkz]

Homework Statement

A cat and a baby are launched from skeet traps on opposite sides of the
Grand Canyon, which are at the same height and a distance ℓ apart. The
cat is launched horizontally at speed (v0/√2), the baby at initial speed v0 at
an angle of (pi/4) above the horizontal. By amazing good luck, it happens that
the cat and baby collide in midair.

For a collision to occur, the cat and the baby must have been launched at different times.

determine the time difference between the launching of the cat and the launching of the baby.

Homework Equations

x(t)=v0*cos(θ)*t
y(t)=v0*sin(θ)*t-[gt^2]/2

The Attempt at a Solution

I set up the equations so that h is equal for both the baby and the cat, with the baby flying for t+tprime seconds, and the cat flying for t seconds. However, this left me with a long, unsimplifiable equation. Next I tried to find h, the height at which they collide, and use that to find the distance L where they do collide. However, this left me with the problem of the gravity, which I couldn't resolve. If anyone could give me a direction of where to go, or how to set up the equations themselves, that is all I really need, and maybe even a general direction of how to proceed with the problem.

 

[Delphi51]

What a problem! At first reading, there seem to be way too many variables, but we must think of Vo and L as being known givens. From them, can we calculate a t' that results in a collision? When you write your Xcat = Xbaby and Ycat = Ybaby equations, you are sure on the write track. You have two equations and two unknowns, t and t'. So you can solve for t'. It is a little messy (use a large sheet of paper), but it works out without even a quadratic. In the horizontal part, did you get t = L*.707/Vo - t'/2 ? In the vertical part I have a t² term, a (t+t') term and a (t+t')² term. After subbing one into the other and squaring the binomials, I got a mess all right - 8 terms - but four of them canceled out, including the (t')² ones.

I scarcely believed my answer, so I made a spreadsheet with the x,y trajectories for both objects, given values of L, and Vo. I had it calculate the t' from L,g and Vo, and sure enough, that resulted in a collision. Slight changes in t' and there was no collision. Amazing!

 

[omoplata]

OK, I had a crack at this and I think I solved it.

This is what I did ( it seems similar to your first approach, so maybe you made an algebraic mistake? ). Let's call the time of flight of the cat t_b and the time of flight of the baby t_b. I found x_c and x_b, which are the horizontal distances traveled by the cat and the baby respectively. Then I found the y_c and y_b which are the distances the cat and the baby respectively fell, and made them equal. This allowed me to find t_c in terms of t_b. Then I got another equation using the fact that x_c + x_b = l , and plugged the expression for t_c into it. The only unknown quantity remaining was t_b. So I solved for t_b and found it in terms of known quantities. Then I found t_c, and subtracted it from t_b to find t_b - t_c.

Edit: Hi Delphi51. I was actually writing out my reply when you posted, so I didn't know you posted first. Yeah, the quadratic terms canceled out in my solution too.

 

[chimpfunkz]

Let me see if I am on the right track:

$$y_b = v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt_b$$

and

$$y_c = \frac{v_0}{√2}sin(0)t_b-\frac{1}{2}gt_c$$

as well as

$$ℓ = x_b + x_c$$ or $$ℓ = v_0 cos(\frac {π} {4}) t_b + \frac{v_0}{√2}cos(0)t_c$$
So when I solve for $$t_c$$ using the formula for length to solve for $$t_c$$, I get

$$t_c=\frac{ ℓ-v_0cos(\frac {π}{4})t_b} {\frac{v_0}{√2}cos(0)}$$

So what do I do from here? do I just plug in $$t_c$$ for the y positions of the baby and cat when they equal each other? or is there something I am missing,

 

[omoplata]

Yes. You just have to plug in $t_c$ to $y_{b} = y_{c}$, but simplifying the expression for $t_c$ will help a lot.

There seem to be mistakes in your $y_b$ and $y_b$, which I guess are absent minded algebraic mistakes, since you have the equation for y(t) correct in your original post.

 

[chimpfunkz]

whoops! I forgot my squares! so the correct equations
$$y_b = v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt^2_b$$ and $$y_c = \frac{v_0}{√2}sin(0)t_b-\frac{1}{2}gt^2_c$$

But how could you simplify tc? would the simplest form just be

$$t_c=\frac{√2( ℓ-v_0cos(\frac {π}{4})t_b)} {v_0}$$

Then after this is where I get lost. I understand the concept, that is to say replacing tc with the above, but whenever I do it out on paper, I seem to get lost, or the terms don't really work well. This might be a little too much to ask, but could you show the first few steps of plugging in tc into y(t)?

 

[omoplata]

You can simplify further using $$\cos \left( \frac{\pi}{4} \right) = \sin \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$$.

Everything else is the same as I did so far.

If you still have trouble I'll post how to plug in $t_c$.

 

[chimpfunkz]

So for Tc, $$t_c=\frac{√2 ℓ-v_0t_b)} {v_0}$$

So you plug that in for when y(t) is equal for both the baby and the cat, which is

$$\frac{v_0}{√2}sin(0)t_c-\frac{1}{2}gt_c= v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt^2_b$$ or, since sin(0)=0,

$$-\frac{1}{2}gt_c= v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt^2_b$$

so plugging in Tc, you get

$$-\frac{1}{2}g(\frac{√2 ℓ-v_0t_b)} {v_0})^2= v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt^2_b$$ then you simplify this to get tb in terms of velocity and gravity right?

 

[Delphi51]

It is actually tb-tc (what you originally called t') that you want.
Recommend you replace tc with tb+t' so you can solve for t' - easier than trying to solve for tb-tc.

 

[omoplata]

@chimpfunkz,

Yes, you simplify that equation to get $t_b$ in terms of velocity and gravity. But like Delphi51 said, the final answer they ask for is $t_b - t_c$. After finding $t_b$, you can plug that into your equation for $t_c$ to find $t_c$. Then subtract from $t_b$ to find $t_b - t_c$.

 

[chimpfunkz]

Alright, so at long last, this is what I have:

$$t_b-t_c=2\frac{\frac{-gℓ}{v^2}}{ \frac{v^2-2gℓ}{√2v}} - \frac {√2ℓ}{v}$$
The question is, what does this simplify too? What should I do to make the algebra a little easier? Would the final result of this match the answer my teacher gave for this problem, which is $$\frac{\frac{ℓ√2}{v}}{\frac{2gℓ}{v^2}-1}$$

 

[omoplata]

Well, it would be if
$$t_b - t_c = 2 \frac{\frac{-gl^2}{v^2}}{\frac{v^2-2gl}{\sqrt{2}v}} - \frac{\sqrt{2}l}{v}$$
instead of what you've written,
$$t_b - t_c = 2 \frac{\frac{-gl}{v^2}}{\frac{v^2-2gl}{\sqrt{2}v}} - \frac{\sqrt{2}l}{v}$$
I had a solution on paper yesterday, but now I've lost it. Check your calculations again, and if you can't find anything wrong, I'll solve it from the beginning and see if my solution is equal to your teacher's.

 

[chimpfunkz]

You were right, I did lose the square. It is actually $$t_b - t_c = 2 \frac{\frac{-gl^2}{v^2}}{\frac{v^2-2gl}{\sqrt{2}v}} - \frac{\sqrt{2}l}{v}$$

And I know this might be a lot to ask, but could you take the formula above, and maybe explain quickly how to simplify it? Algebra is not really my strong suit.

 

[omoplata]

OK, here are the steps.
$$t_b - t_c = 2 \frac{\frac{-gl^2}{v^2}}{\frac{v^2-2gl}{\sqrt{2}v}} - \frac{\sqrt{2}l}{v}$$
First let's take this to be the subtraction of two fractions.
$$t_b - t_c = \frac{\frac{-2gl^2}{v^2}}{\frac{v^2-2gl}{\sqrt{2}v}} - \frac{\sqrt{2}l}{v}$$
If you look at your teacher's answer, it's one fraction with only one denominator, namely $\frac{2gl}{v^2} - 1$. So let's try to get one denominator for our expression. We can do this by multiplying both the numerator and the denominator of the left fraction by the denominator of the right fraction, and multiplying both the numerator and the denominator of the right fraction by the denominator of the left fraction. That is,
$$t_b - t_c = \frac{\frac{-2gl^2}{v^2}}{\frac{v^2-2gl}{\sqrt{2}v}} - \frac{\sqrt{2}l}{v} = \frac{v \frac{-2gl^2}{v^2}}{v \frac{v^2-2gl}{\sqrt{2}v}} - \frac{\frac{v^2-2gl}{\sqrt{2}v} \sqrt{2}l}{\frac{v^2-2gl}{\sqrt{2}v} v} = \frac{\frac{-2gl^2}{v}}{\frac{v^2-2gl}{\sqrt{2}}} - \frac{\frac{v^2l-2gl^2}{v} }{\frac{v^2-2gl}{\sqrt{2}}} = \frac{\frac{-2gl^2}{v} - \frac{v^2l-2gl^2}{v}}{\frac{v^2-2gl}{\sqrt{2}}}$$
So,
$$t_b - t_c = \frac{\frac{-2gl^2-v^2l+2gl^2}{v}}{\frac{v^2-2gl}{\sqrt{2}}} = \frac{\frac{-v^2l}{v}}{\frac{v^2-2gl}{\sqrt{2}}} = \frac{-\sqrt{2}vl}{v^2-2gl} = \frac{\frac{l\sqrt{2}}{v}}{\frac{2gl}{v^2}-1}$$
which is your teacher's answer. If you have trouble following this, let me know where and I'll elaborate.

 

[chimpfunkz]

Wow, that makes it so much clearer. The only step I don't get is how you get from $$\frac{-\sqrt{2}vl}{v^2-2gl}$$ to $$\frac{\frac{l\sqrt{2}}{v}}{\frac{2gl}{v^2}-1}$$

 

[omoplata]

Multiply both numerator and denominator by -1.

Then divide both by $v^2$.