Solving Projectile Problem for Initial Speed v Needed

[tony873004]

*** edit to fix a few mistakes pointed out by Davorak in the 2nd post to this thread ***

A cannon is used to launch capsules of fire retardant chemical onto a fire. The cannon is 1.0 km from a cliff that is 0.5 km high. The fire is 1.0 km from the base of the cliff. If the capsules are launched at an angle of 45 degrees, what initial speed v is necessary?

List what I know:
The cannon is 500 meters above the height of the fire.
The cannon is 2000 meters from the fire in the x direction.

v_{iy}=sin45*v_{i}
v_{ix}=cos45*v_{i}

Now try to solve it:

t_{x} = t_{y}

t_{x} = \frac{2000}{v_{ix}}

Using the formula:
y_{f} = {y_{i}+v_{iy}t-\frac{1}{2}gt^2=0

Construct a quadratic equation:
t_{y} = \frac{-v_{iy} \pm \sqrt{v_{iy}^2-4*4.9*yi}}{2(-\frac{1}{2}g)}

Set it equal to t_{x}

\frac{2000}{v_{ix}} = \frac{-v_{iy} \pm \sqrt{v_{iy}^2-4*4.9*yi}}{-g}

Substitute my values for v_{iy} and v_{ix} and y_{i}

\frac{2000}{cos45*v_{i} } = \frac{-(sin45*v_{i}) \pm \sqrt{(sin45*v_{i})^2-4*4.9*500}}{-9.8}

Now if I could only knew how to extract v_{i} out of this mess, I'd have my answer.

How do do that, or...

Is there an easier way to tackle this problem?

 

[Davorak]

t_{x} = \frac{v_{ix}}{2000}

Hmm this would be.
t_{x} = \frac{v_{ix}}{d_x}
so that would mean?
t_{x} d_x = v_{ix}
Remember v*t = D

<br /> t_{y} \neq{v_{iy}+v_{iy}t-\frac{1}{2}gt^2 \neq 0 <br />
<br /> {x_{iy}+v_{iy}t+\frac{1}{2}gt_y^2 =x_{yf}<br />
What does x_{yf} equal?

 

[tony873004]

Hmm this would be.
t_{x} = \frac{v_{ix}}{d_x}
so that would mean?
t_{x} d_x = v_{ix}
Remember v*t = D

<br /> t_{y} \neq{v_{iy}+v_{iy}t-\frac{1}{2}gt^2 \neq 0 <br />
<br /> {x_{iy}+v_{iy}t+\frac{1}{2}gt_y^2 =x_{yf}<br />
What does x_{yf} equal?

Thank you for catching that. I knew with all the TEX I'd slip. It should read
x_{f} = {v_{iy}+v_{iy}t-\frac{1}{2}gt^2=0

But I don't think the error followed me to my next line, where I made a quadratic equation out of it.

And you're right. I inverted my v and d. But that will only invert the left side of my equation and leave me with the same problem.

Thanks for catching those. I'm going to edit the post. Any ideas on how to solve it.

 

[Davorak]

The t = \frac{D}{v} followed again a few lines down.

\frac{2000}{v_{ix}} = \frac{-v_{iy} \pm \sqrt{v_{iy}^2-4*4.9*yi}}{-g}

Also your corret equation is still wrong:

x_{f} = {v_{iy}+v_{iy}t-\frac{1}{2}gt^2=0

The units of this would be
[displacement] = [velocity]+[displacement]+[displacement]

x_{f} \neq {v_{iy}+v_{iy}t-\frac{1}{2}gt^2 \neq 0

The final x_{fy} position does not equal zero that would mean that it did not move at all.

This is the equation I think you are getting at:
{x_{iy}+v_{iy}t+\frac{1}{2}gt_y^2 =x_{yf}

 

[tony873004]

t = \frac{D}{v}
I changed that one too. I believe it is in the correct order after my last edit.

the line
x_{f} = {v_{iy}+v_{iy}t-\frac{1}{2}gt^2=0

should have read
y_{f} = {y_{i}+v_{iy}t-\frac{1}{2}gt^2=0

Thanks for catching that. I believe it was just a typo though, as I don't believe it follows me into my next line of the quadratic equation.

I got this equation by rewriting one in the book
x=x_{0}+v_{0}t-\frac{1}{2}gt^2=0

Thank you for your help so far

 

[Davorak]

No problem on the helping part. I am pretty sure though that:
y_{f} = {y_{i}+v_{iy}t-\frac{1}{2}gt^2=0
will not equal zero. This is becasue y_{f} can not equal zero when the chemical retardant reaches the fire. The cannon is 500 metters above the house.
"The cannon is 500 meters above the height of the fire."
That means y_f = -500 \ {When \ g \ is \ considered \ negative \ or \ } y_f = -500 \ {when \ g \ is \ considered \ posative}

 

[tony873004]

But y initial, the height of the cannon was considered 500m. I didn't use 500 until I started plugging into my quadratic. But if the cannon is 500 then the ground should be 0. Right??

 

[Davorak]

That is how I would do it. It ussaly simplifes things to consider the starting position as the zero point. You do not have to do this if you are unconfortable with it.

more to come in a moment

edit:
ahh yes I did not notice your edit until now.

 

[tony873004]

That is how I would do it. It ussaly simplifes things to consider the starting position as the zero point. You do not have to do this if you are unconfortable with it.

more to come in a moment

edit:
ahh yes I did not notice your edit until now.

I'm copying an example in the book where a balloon is rising at a constant speed, and someone drops something. The balloon is 20m high at the time, and the example uses 20 as y initial, and 0 as the ground. More to come?...

 

[Davorak]

These types of problems are usually easier to solve if you break of the y component into two parts.

1. Find the time and the displacement for where the projectile reaches the top of it's arc. The top of the arc is where a = 0 so finding the time it takes is easy. Once you know the time it is simple to find the height change.

2. Find the time it takes for the projectile to drop from its maximum height to its final height. This is simpler since the equation becomes:
x_f = x_i-\frac{1}{2}g t^2

Breaking the problem up this way adds steps but each step is much simpler then the one you are trying to solve.

does this help?

 

[tony873004]

I agree with you that it's easier to break it up into 2 steps if time was what I was after, and I was given either an initial velocity and an angle, or the y-component of the initial velocity.

But the tricky part of this problem comes from the fact that initial velocity is not given in any form, and velocity is what we must compute.

That's why I constructed it as a simultaneous equation solution. But I don't know how to split half of a simultaneous equation into 2 parts, so I used the longer method that involves creating a quadratic equation so it can be solved all in one step. Also, I'm pretty sure that in the +/- part of the quadratic, that I'm going to discard the + and keep the -.

I appreciate your help. We get 12 problems on our homework, but he only grades one problem. Do 1 problem, but do the one he grades give you a +, but do 11 problems and skip the one he grades gives you a 0. And he told us today that the graded problem would be a projectile problem. This is one of 3 projectile problems, and the only problem out of the 12 that is giving me problems.

 

[Davorak]

blah... I thought of a better way find the time on the x-axis then plug that into the equation for y distance.

<br /> t_x = \frac{2000}{v_{ix}}<br />

<br /> x_f = x_0 +v_{iy} (\frac{2000}{v_{ix}}) -\frac{1}{2}g(\frac{2000}{v_{ix}})^2<br />

viy=vix=vi

solve for vi

 

[tony873004]

blah... I thought of a better way find the time on the x-axis then plug that into the equation for y distance.

<br /> t_x = \frac{2000}{v_{ix}}<br />

<br /> x_f = x_0 +v_{iy} (\frac{2000}{v_{ix}}) -\frac{1}{2}g(\frac{2000}{v_{ix}})^2<br />

viy=vix=vi

solve for vi

viy^2 +vix^2 = vi^2

so I could re-write that as: (change the Viy to Vix since they equal each other)
x_f = x_0 +v_{ix} (\frac{2000}{v_{ix}}) -\frac{1}{2}g(\frac{2000}{v_{ix}})^2

give me a few minutes to look this over.

 

[tony873004]

since x_{0}=0 and in v_{ix}(\frac{2000}{v_{ix}})the v_{ix} 's should cancel, and x_{f}=2000, I'm left with:
2000 = 2000 - \frac{1}{2}g(\frac{2000}{v_{ix}})^2

but only 2000 - 0 = 2000, and infinity is the only number I can plug in for Vix to make everything to the right of the minus sign equal 0.

Did I do this right? Algebra is not my strong point

 

[Davorak]

i was puting xf and xi but I ment them only in general terms they are yf yi. Sorry for not using consisant notation. yi = 500 yf =0 with the way you have the problem set up. So:
0= 500 +2000 - \frac{1}{2}g(\frac{2000}{v_{ix}})^2

 

[tony873004]

If I did my algebra right, vix comes out to 88.5, which seems reasonable. Stay tuned... I'm going to try this a different way and see if they're the same. back in 15m!

 

[Davorak]

Easiest way to cheak is to plug the numbers back into the equation. Find the time taken for the projectile to get to the house and then plug the time and the velociy in the y euation to to make sure to gives the projectile reaching zero meters.

 

[tony873004]

Easiest way to cheak is to plug the numbers back into the equation. Find the time taken for the projectile to get to the house and then plug the time and the velociy in the y euation to to make sure to gives the projectile reaching zero meters.

That makes sense. Let me try that. I'll show you my more complicated method of checking in a few minutes.

 

[tony873004]

Your method of checking seems to work. I get 22.58 seconds now that backwards checking is simple kinmatics. 22.58 * 88.5 = 1998.33, close enough to 2000 to blame on rounding errors.

Thanks. Sometimes I check these kind of things by writing simple computer programs to numerically integrate the problem and break it into a few thousand steps, updating x,y, and vy using g with each step. In this case, I would have started plugging in guesses for Vi, and zeroing in on it until my x value equaled 2000. But I don't have to do that now. I didn't quite follow the step where you came up with the magic formula that simplified the whole thing
0= 500 +2000 - \frac{1}{2}g(\frac{2000}{v_{ix}})^2

 

[Davorak]

Ok no magic here
<br /> t_x = \frac{d}{v_x}<br />
and
<br /> x_{fy} = x_{iy} +v_{iy} t_y -\frac{1}{2}g*t_y^2<br />
But we know that t_y = t_x the projectile can only take one time to get to the house so they must be equal to each other. so
<br /> x_{fy} = x_{iy} +v_{iy} t_y -\frac{1}{2}g*t_y^2 = x_{iy} +v_{iy} t_x -\frac{1}{2}g*t_x^2<br />
but we know:
<br /> t_x = \frac{d}{v_x}<br />
Therefore:
<br /> x_{fy} = x_{iy} +v_{iy} t_y -\frac{1}{2}g*t_y^2 = x_{iy} +v_{iy} (\frac{d}{v_x}) -\frac{1}{2}g*(\frac{d}{v_x})^2<br />
Where d is the horizontal displacement
Does that make sence?

 

[tony873004]

That makes sense. It looks like a simultaneous equation, but easier than the one I came up with.

Thanks for all your help. This isn't due till wednesday, so if I have any more questions, I'll ask tommorow.

 

[tony873004]

Just as a triple check, this simple vb 6.0 program let's you keep guessing at initial velocity by entering values into text5 until your x value = 2000 as displayed in text8. And the results agree with your formulas.!

Private Sub Command1_Click()
Dim theta As Double
theta = Val(Text1.Text) 'initial angle in degrees of projectile at launch
ch = Val(Text2.Text) ' cliff height = 500
cc = Val(Text3.Text) ' cannon to cliff = 1000
fc = Val(Text4.Text) 'fire to cliff
vi = Val(Text5.Text) 'initial velocity

xv = vi * Cos(radians(theta))
xiy = vi * Sin(radians(theta))
vy = xiy
ts = 0.0001 'time step
g = 9.8
xi = 0
yi = ch

x = xi
y = yi
While y > 0
    y = y + vy * ts
    x = x + xv * ts
    vy = vy - (9.8 * ts)
    DoEvents
Wend
Text8.Text = x
End Sub

Function radians(degrees As Double) As Double
    pi = Atn(1) * 4
    radians = (degrees / 180) * pi
End Function