# Happy g day

Actually, it should be called Happy "Magnitude of the local Earth gravitational field" day. You know, 9.8 N/kg on September 8 (9/8). Get it? Well, the idea was for the physics students and faculty to build some stuff to do outside - projectile motion type stuff. Well, we had the idea a while ago and then kind of forgot about it.

In order to just get something done, I set up the "shoot the falling target" demo. (previously known as shoot the monkey). Here is a quick video demo (seriously - first take too).

What is going on here and what does this have to do with g? Well, it doesn't really have anything to do with g, but it is a cool demo (meaning this demo would still work on the moon). The basic idea is that you aim a projectile launcher straight at a hanging target. When the ball is launched, the target drops and the ball collides with the target in mid-air.

I guess it is clear, but the vector v0 is the initial velocity of the ball aimed at an angle θ at a target that is a distance s horizontally away and a height h above the launcher. But why do they hit? Let me start with the ball. (here is a quick projectile motion review) The ball's motion can be broken into an x-direction motion and a y-direction motion with the following initial velocities:

Two quick notes: I am using v0 to represent the magnitude of the initial launch speed. Also, the x-velocity does not change (no forces in the x-direction) so that I can just call it vx instead of vx0 (of course, the y-velocity does change). Using those velocities, I can get position-time functions for both directions.

There is the "g"! Also, I am assuming the ball launches at time t= 0 seconds from a position x0 and y0. Now, what about the target? The target also starts falling at t = 0 seconds, but its initial position is (x0 + s,y0 + h) as seen in the diagram. There is no x-direction motion for the target, so I can just write:

I am calling ym is the y position of the target as a function of time. The target's initial velocity is zero. First question: what time will the ball be at the x-coordinate of the target? Going back to the function for the x-position of the ball,

And what will be the vertical position of the ball at this same time?

What about the target? Where will the target be at this same time?

Notice that the only difference between the y position of the target and the ball is that the ball has a "s tan(θ)" where the target just has an h term. Oh but wait. Look back at the original diagram and you will see that:

What does this mean? This means that at the time when the ball reaches the x-position of the target the ball and the target have the same y-position. If two things are at the same place at the same time, they hit. Boom.

Notice one cool thing - the value of the magnitude of the initial velocity does not matter (unless the ball hits the floor before it gets to the target). The only important thing is that the ball is aimed at the target. This means that θ for the velocity vector is the same θ for the geometrical set up of the target.

Wait! There is more! How about a nice python-generated graph? This is pretty simple to set up really. In python, I will model the shot ball and the falling target.

That is with a launch speed of 6.5 m/s. What if I increase the speed to 8 m/s?

You can't really see that the two paths cross at the same time. Let me draw fewer points.

There, that is with a time step of 0.1 seconds. The points are numbered and you can see that the two points labeled 4 are right next to each other.

That is it for g-day. I will have bigger plans for next year.

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I also love the shoot-the-monkey demo, and your demo of it was well executed. The huge arc the projectile takes (and hitting just before it falls out of frame) adds to the drama.

There should be some simple way to explain this in basic principles without any trigonometry, no?

Like earth's gravity causes constant acceleration, so what you see in your earth-based lab would have to match what would happen in a zero-g lab as observed by someone in a neighboring spaceship that's accelerating. Which then makes it pretty obvious that if the launcher is pointed at the monkey...

Or something about the linearity of something or other?

By Anonymous Coward (not verified) on 08 Sep 2010 #permalink

@AC,

I guess you could describe this in the accelerating frame of the falling objects. In that case, the target would stay at the same relative position with respect to the launched ball.