[Izzy]

The.. numbers.. don't work? Which makes no sense, because they should.

We were launching rockets, the following was yielded.

Launch angle: 50°

Range: 244 ft

Time: 3.39 s

So, I'm supposed to find the initial speed, initial horizontal velocity, and initial vertical velocity. Easy enough. Since Δy = 0, let's use the range equation:

R = ((vi^2)(sin(2theta)/g)

Rearrange for vi:

SQRT((R*g)/(sin2theta)) = vi

Since we're using feet as units, use 32 for gravity (just magnitude), and..

SQRT((244*32)/(sin2*50)) = 89.0 ft/s

Good enough so far. Let's break it down into vertical and horizontal components.

89.0cos(50) = x = 52.7 ft/s x-hat

89.0sin(50) = y = 68.2 ft/s y-hat

Cool, it looks solved, right?

I decided to check my work with Δx = v_x*t

...

Δx = 52.7 ft/s * 3.39 s

244 ft ≠ 179 ft

D:

[Guest]

What you are seeing here is that the mathematical model used does not accurately describe the real world to a high precision.

When the equation "R = ((vi^2)(sin(2theta)/g)" was derived, many simplifying assumptions were made.

For example, it was assumed that air resistance is negligible, that gravitational acceleration varies negligibly with height, that there was no wind (or any atmosphere at all), ... etc.

Making these assumptions is a trade off, because if you were to take into account all of those things, calculations would be much more difficult.

In the practical world, cost and effort are weighted against correctness. Finding the right balance is usually finding a mathematical model that is "correct enough".

More accuracy means more effort and more cost.

So do not panic, the laws of physics aren't being broken... the only problem here is that the equations you used didn't account for some other factor significant enough to create such a difference.

If you are doing a report or something, you might want to identify some of potential factors to hypothesize why the measurements aren't entirely consistent with the calculations.

Edited October 12, 2010 by mmiguel1

[bonanova]

Izzy, if you want to just consider the geometry of this experiment,

try writing all the simple relationships you know among the variables stated in the problem.

Initial speed, launch angle, v_x, v_y, range, time.

Then look for a relationship where only one of the variables is unknown.

Plug that result into the other relationships and repeat.

.

1. You know the relation of v_x to range and time.
2. You know the relation of v_y to launch angle and v_x
3. You know the relation of initial speed to v_x and v_y, or to launch angle and v_x or v_y..

.

[Guest]

Izzy, if you want to just consider the geometry of this experiment,

try writing all the simple relationships you know among the variables stated in the problem.

Initial speed, launch angle, v_x, v_y, range, time.

Then look for a relationship where only one of the variables is unknown.

Plug that result into the other relationships and repeat.

.

1. You know the relation of v_x to range and time.
2. You know the relation of v_y to launch angle and v_x
3. You know the relation of initial speed to v_x and v_y, or to launch angle and v_x or v_y..

.

I don't think that the observed data satisfies the theoretical equation.

If I drop a ball from 4.8 meters above the ground, theory says it should take sqrt(2y/g) or about 1 second.

If I measure 2.46 seconds, then I will be confused. According to high-school textbook theory, y = gt^2/2 if the ball fell for 2.46 seconds, then the ball should have been at a height of 11.025 meters. But I was careful to drop it from 4.8 meters.

y = gt^2/2 is idealized and does not account for some physical factors that change the answer.

If I do lots of research and come up with a better equation that takes into account these physical factors, then I might calculate 2.46 seconds.

In my example, let's say for example that I was actually doing the experiment on the moon, and g is much different on the moon than on earth. I calculate now that on the moon, it should actually take 2.46 seconds to drop a ball 4.8 meters.

By changing g, I make my theory more match the physical situation I am in.

Of course, your rocket isn't on the moon Izzy, so the fix to your equation is not as simple as changing a constant.

It's sort of disappointing, but basically all mathematical models are estimations of physical reality. You know you have a very good theory if the mean of your observations is the theory predicted value, and measurements don't deviate from that value very much (the variance is low.

Your measurements, if done correctly are right. It is the equation that has failed here... not to say it is a bad equation. Just look at it as an estimation (like all other things when math is applied to the real world). It is most likely the best estimator you are expected to use for this experiment. If you want more accuracy, you can look up more advanced models for rocket trajectories on google or something.

Edited October 12, 2010 by mmiguel1

[Izzy]

Again, thanks for the help, or rather, reassurance. Though, I did use the wrong angle (the above was from the vertical, and I meant to use the horizontal), so the difference between the two methods ended up being only 10ish instead of the kind of scary 65. I had time to fix it though, so it wasn't a problem.

[Guest]

Again, thanks for the help, or rather, reassurance. Though, I did use the wrong angle (the above was from the vertical, and I meant to use the horizontal), so the difference between the two methods ended up being only 10ish instead of the kind of scary 65. I had time to fix it though, so it wasn't a problem.

One of the important things I think for your rocket is that it continues to be accelerated after it has left the ground.

The trajectory equation used assumes that all non-gravitational acceleration occurs at the moment of launch (like a cannonball).

Unlike a cannon-ball, your rocket continues to burn fuel as it cruises through the air, and hence continues to accelerate and go further than a cannonball would with the same initial velocity and position.

It just occurred to me, and I thought I might as well mention it.