Projectile Motion Question - Clearing a Wall

  • Thread starter Vyl
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  • #1
Vyl
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Homework Statement



Among many other things, determine if a projectile will clear a wall. I am attempting to make a program to respond with a 'yes' or 'no'. So this is not a specific question, it is more of a general "how is this solved?" question.

The known values are initial height, initial angle, and initial velocity.

The outputs are the maximum height, time, and distance, which I have calculated and outputted correctly using equations in the textbook.

So my basic question is, I need to figure out that given a distance from a wall and height of the wall, will the projectile clear it?

Homework Equations



dMaxHeight = dInitialHeight + (dInitialVelocity * dInitialVelocity * sin(dInitialAngle) * sin(dInitialAngle)) / (2 * dGravitationalAcceleration);

dMaxTime = (dInitialVelocity * sin(dInitialAngle)) / dGravitationalAcceleration + sqrt((2 * dMaxHeight) / dGravitationalAcceleration);

dMaxDistance = dInitialVelocity * dMaxTime * cos(dInitialAngle);


I believe that these equations are correct as I have checked them with the Java simulations.

The Attempt at a Solution



I've sketched out the parabola and the wall, and the only thing that I can think of is to make a chart of the x and y values of the travelling object as time progresses, and test them with the height and y-coordinate of the wall. Is this even close to being correct? I hope I'm being clear of what I'm trying to figure out.

Thanks very much for your help!

- Vyl
 

Answers and Replies

  • #2
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Yes, I believe your attempt is on the right track. Find Height as a function of horizontal distance, evaluate height at the place where the wall is, see if the height is greater than the height of the wall.
 
  • #3
Vyl
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Problem Solved

Yes, I believe your attempt is on the right track. Find Height as a function of horizontal distance, evaluate height at the place where the wall is, see if the height is greater than the height of the wall.

Thank you for the reply!

I believe that I have figured it out -- I did exactly as you said. Thanks for the help!

- Vyl
 
  • #4
Vyl
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A Little More Help

Yes, I believe your attempt is on the right track. Find Height as a function of horizontal distance, evaluate height at the place where the wall is, see if the height is greater than the height of the wall.

Hi,

Well ... I was able to get a working model (I think it works) in C++, but for some reason I can't get the Excel spreadsheet to work properly. May I have a little more clarification please? I created a table of X and Y values for the entire length of time that the projectile is in the air. What I don't know how to do is find if (and where) the wall will meet the projectile path.

How do I evaulate height at the place where the wall is? For my sample data, I know that the wall is 4 meters out and 4 meters high (this data should make the projectile collide). Should I run through my X and Y values, and when I hit 4 in the X side of the table, look at the cooresponding Y value, and compare it with the height of the wall? That sounds like it would work ... I'm not sure how to implement it in Excel though.

Any hints or guidance would be greatly appreciated. Thanks very much!

- Vyl
 
  • #5
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If you are able to calculate a table of X and Y values, can you not also calculate the Y value for a given X, namely for X = horizontal wall distance.

If you are depending on time, you need to do as nicktacik says and eliminate time, finding an equation relating height of the projectile with its horizontal distance traveled.

Because the horizontal velocity stays the same, the equation for horizontal distance (x) traveled is:

[tex]x = v_0\cos\theta \cdot t[/tex]

where v0 is the initial velocity, [tex]\theta[/tex] is the launch angle, and t is the time.

For the vertical movement, you use the constant acceleration equations with g = -9.81 m/s^2 for your acceleration:

[tex]y = y_0 + v_0\sin\theta \cdot t - \frac{1}{2}gt^2[/tex]

Now, all you need to do is eliminate t.
 
Last edited:

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