# Projectile Motion.

## Homework Statement

I have this lab and my goal is to find how far I should place a box if I was to catch a plastic leprechaun that weighs 22.4 grams fired out of an air cannon. I did this lab and the final velocity I ended with was 12.76 meters/sec (29.72 meters was the average distance and 2.33 seconds was the average time)
Ignore air resistance.

## Homework Equations

d=vit+(1/2)at2 was ah equation I remember using while first doing projectile motion in class but because I cannot find my notes I am stumped.

## The Attempt at a Solution

d=?
vi=0m/s
t=2.33 sec
a=9.8 m/s2(gravity)

when I plugged all those values in the equation I gave I ended up with d=26.06 meters.

Now would this be right since my average distance was 29.72 meters?

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It seems like you are saying you ignored air resistance when you did the lab. That wouldn't really be possible, so the experimental data you collected does include air resistance. Then when you're solving it theoretically, I'm assuming you know the initial velocity of the gnome from the cannon, and maybe an angle?

That is the right equation, you have vi set equal to zero, but if it was shot out of an air cannon, how can that be right?

So the point is, I'm not sure exactly what you are doing, but to solve for the x-displacement (d) of a projectile, you only need d=vixt. When you ignore drag there is not acceleration in the x direction, so that initial velocity is constant.

Good luck, and if you can provide some clarification then we can help more.

The way I was taught my teacher told me that to find the distance an object will travel horizontal can be deduced by d=vit+(1/2)at2. She said this is right because the initial velocity of an object will always be zero since before it is shot, thrown, or whatever it is at rest. She said gravity than always brings back everything back down at the same rate when we ignore air resistance (yes, I agree it's stupid to do so but whatever) and that if we have the time of flight than we can find the distance it was launched, thrown. So yes my data will be completely unrelated to my actual answer but what I was wondering was, does this work?

Unless you are also going to consider forces, you can't say the initial velocity of the object is 0. Perhaps though, the cannon fires horizontally from an elevated position, so all of the initial velocity is on the x axis, then the initial y velocity would be 0. If this is the case, then you can solve for the height because you know the time.

One of the biggest problems people have when using kinematics equations is that they don't ensure all of the vectors involved are in the same dimension.
This means that when you are solving for d, a displacement on the x axis, that vi must be the x component of the velocity, and that a must be the acceleration on the x axis (there is none however)

You can write that equation for both y and x motion. When you write it for x, you have d=vixt and when you write it for y, you have h=1/2(g)t^2 assuming the initial velocity in the y direction is 0.

I don't know exactly what your variables are, but often times problems will require both the x and y equations, and they can be combined through the time variable.

And finally, the bit about the initial velocity being 0. I definitely see why you think that, but if the problem is dealing only with kinematics you should probably consider vi to be the muzzle velocity of the cannon. Otherwise you need to know the acceleration that the air cannon imparts on the projectile, and for how long. Then you could find the initial velocity as it leaves the cannon. I doubt the problem would be setup this way unless you are already working with forces.

I hope some of this makes sense, I'll try to better explain if something doesn't.