# Projectile Trajectory (I'm confused about the lab report)

1. Jan 30, 2010

### bedizzy

The experiment was to measure muzzle velocity of a ball fired by a 'spring gun'. Then to use the equations of projectile motion in order to determine the projectile trajectory. During the experiment we used a variable-height platform to stop the ball mid-air (at y = 0.4 m & y = 0.1 m). Now we're suppose to use the data to determine the final landing point of the trajectory. I have all my data, and I've created the projectile trajectory in excel from the initial position to y = 0.10 m.

Now I'm lost on how to complete the trajectory!? The professor wants us to use the quadratic equation to solve for Xp (Xp is the final point of impact, where y = 0). But I don't know what information to plug into the quadratic equation. And even if I did, it doesn't make sense to me because the range should just be v0 * t.

Can somebody please nudge me in the right direction? If I'm not explaining this clearly, please let me know and I'll post my lab data sheet. I just need to know how to complete this trajectory using the quadratic equation to solve for xp.

Last edited: Jan 30, 2010
2. Jan 30, 2010

### !kx!

I couldn't understand the problem very well. I think that with y, you mean the height above ground.. And that should make x along the ground.. what is X0??

3. Jan 30, 2010

### bedizzy

sorry, x0 is suppose to be v0.... I just corrected it...

And yes, Y is the height above the ground. The initial y position was 1.20 m. After the projectile was fired, we used a platform to stop the ball at ya = 0.4 m, and yb = 0.1 m. Using experimental data and projectile motion equations I determined xa and xb.

When the ball hits the ground I know that y = 0, but I don't know how to find the x position (Xp).

The professor wants us to show the final trajectory on an excel chart, and told us to use the quadratic equation to solve for xp.

I attached the diagram from my lab report.

#### Attached Files:

• ###### apparatus.JPG
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4. Jan 30, 2010

### !kx!

The problem is pretty straight forward.. First of all you should know about the quadratic equation you are talking about..
Your diagram gives you 2 points on the curve.. that should help you find the constants in your quadratic equation.. After you get the equation in complete form, with only x,y variables, substitute the last point (y=0,x=?) to get the result..

5. Jan 30, 2010

### !kx!

finding range shouldn't be a problem if you know initial velocity and time taken for trajectory.. but that way, you won't be able to plot the trajectory, as you require.

6. Jan 30, 2010

### bedizzy

I think I got it; the numbers make sense and the chart looks good. I think my problem was that I forgot how to use the slope in the quadratic equation.

Thanks for the help, I really appreciate it...