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

AI Thread Summary
The discussion revolves around a lab report focused on measuring the muzzle velocity of a ball fired from a spring gun and determining its projectile trajectory. The participant is confused about using the quadratic equation to find the final landing point (Xp) after stopping the ball at specific heights. Clarifications were provided regarding the initial height, the use of experimental data, and how to derive the quadratic equation using two points on the trajectory curve. The importance of substituting the final height (y=0) into the completed quadratic equation was emphasized for finding Xp. Ultimately, the participant gained clarity on using the slope in the quadratic equation to complete their analysis.
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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.
 
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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??
please be more clear!
 
!kx! said:
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??
please be more clear!
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.
 

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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..
 
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.
 
!kx! said:
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..


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...
 
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