# Okay, so I don't usually ask for homework help, but...

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Posted (edited)

So we have this project in Physics (a group project), and we built a ballista for it. It's basically a really big crossbow. Except ours is really small, so it's basically a stationary crossbow. But made out of K'nex because we couldn't get ahold of anything super high quality that would bend enough in the limited build size we had, and K'nex is easily modifiable. Skipping all of the fluff, the goal of the experiment was to see whether mounting it on wheels would have any effect on the range (duh, they reduce the range, it should be a no brainier, but apparently it isn't). So we did that experiment, recording launch angle and range.

Here's the problem. Our teacher wants us to calculate maximum height reached by the projectile, as well as launch velocity and stuff. So I plugged the range and angle into an equation to get the maximum height.

But it came out really low. I triple checked myself, and I'm absolutely sure it went more than 1.5 feet upwards... I'd say more than 4 feet. But I think I found out why. The launch actually happens 9 inches above the ground, which throws off our equations because the launch and landing altitudes are different.

With the data I have, I don't think I can calculate the launch velocity. Not with the calculations I know, at least.

What I know:

Without wheels: Range is 9.36 feet, launch angle is 33 degrees, launch elevation is 0.75 feet (yeah, sorry for feet, in our situation it was easier to measure in feet due to the materials available).

With wheels: Range is 8.64 feet, launch angle is 33 degrees, launch elevation is 0.833 feet.

G is 32.2ft/s^2 (probably obvious but I think I should list it here).

Diagram:

With the equations given, I can calculate the launch velocity if the two points were at the same elevation. But they aren't, and I don't know the range at that elevation, only the range at altitude 0. And I can't find that, because I don't know the impact angle, which is steeper than 33 degrees because the projectile had more time to fall. But how much time? I don't know. I could approximate it to 33 or 40 degrees but that wouldn't be accurate. The only way I can think of doing this would be to draw this on something that can do graphs and then manipulate the parabola to to start and end at that point while having a 33 degree angle, and then I'd have the final angle and the maximum height, which I think I can use to find the initial launch velocity, but I don't know how to express that mathematically, which is something that is strongly encouraged.

Thoughts? If this is an impossible problem then I'll alter our paper so that it displays the values assuming the same altitude, but with a disclaimer saying that it's just for a comparison and isn't really accurate.

Edit: I could do several more trials with the ballista and record average time or max height, or even set up a slow motion camera, ruler, and stopwatch to find the launch velocity, but I most likely won't have the time to do that before the project is due.

Edited by Ultimate Steve

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46 minutes ago, Ultimate Steve said:

With the data I have, I don't think I can calculate the launch velocity. Not with the calculations I know, at least.

Velocity is a vector, in your case consisting of a forwards component and an upwards component.

Since there's no forward or backward acceleration, the forwards component is constant, and you can use d = v * T to find it. Just divide the horizontal distance traveled by the total time of flight, done.

The upwards component is almost as easy. There's constant acceleration of -32.2 ft/s/s, so if the initial upwards velocity we're solving for is v0, then at time t it's v(t) = v0 - 32.2 ft/s/s * t. When the projectile reaches the highest point of its parabola, the vertical component of the velocity is 0 (think of this as gravity gradually erasing the initial upward velocity), so if you can determine that time by observation (meaning run your stopwatch till the peak), you can plug it in to solve v0 = 32.2 ft/s/s * ttop. Combine with the horizontal component to get the complete vector; if you need a single scalar value, try sqrt(horizontal * horizontal + vertical * vertical).

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1 minute ago, HebaruSan said:

﻿ Velocity is a vector, in your case consisting of a forwards component and an upwards component.

Since there's no forward or backward acceleration, the forwards component is constant, and you can use d = v * T to find it. Just divide the horizontal distance traveled by the total time of flight, done.

The upwards component is almost as easy. There's constant acceleration of -32.2 ft/s/s, so if the initial upwards velocity we're solving for is v0, then at time t it's v(t) = v0 - 32.2 ft/s/s * t. When the projectile reaches the highest point of its parabola, the vertical component of the velocity is 0 (think of this as gravity gradually erasing the initial upward velocity), so if you can determine that time by observation (meaning run your stopwatch﻿﻿﻿ till the peak), you can plug it in to solve v0 = 32.2 ft/s/s * ttop. Combine with the horizontal component to get the complete vector; if you need a single scalar value, try sqrt(horizontal * horizontal + vertical * vertical).

Alright, thanks. That does sound like the easiest way, but unfortunately that means I have to run a few trials to get an average time and I won't have class before the project is due... I could try getting to school early and doing it, but that will be a challenge given how I don't agree with mornings!

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Posted (edited)

OK, so I took a geometric tactic and I think I got it. Please run the numbers again, I used 2 digits after the decimal point and that wasn't accurate enough (when I graphed it at the end my x-ints were a little bit off), and I may have done some math wrong thanks to how late it is:

First of all, a bit of calculus. Calculus can determine the slope of a graph at any point. This is done by finding the derivative of a function, which is another function that returns a line with the slope for any point on a function. The derivative of ax^2 + bx + c is ax + b.

First of all, we know two things about your trajectory (working from the stationary one here) it passes through (-4.68, 0.75) and (4.68, 0), these being the launch and landing points. Alone, these are not enough to find a specific parabola. However, we also have the launch angle. We convert angle into slope with tan(theta), so tan(33) (make sure to use degrees, not radians) is ~0.65. We know that f(x) = ax^2 + bx + c, and the derivative, f'(x) = ax + b. We know that at x = -4.68, the function has a slope ~0.65. Therefore f'(-4.68) = 0.65 + b. We don't care about b, so we just go for the fact that ax = 0.65, or a(-4.68) = 0.65. We solve to get a ~ -0.14.

Now we have a foot in the door, and with two other points we can find b and c.

f(-4.68) = 0.75
f(4.68) = 0

Using a system of linear equations:

a(-4.68)^2 + b(-4.68) + c = 0.75
a(4.68)^2 + b(4.68) + c =0

This wouldn't be enough to solve, but for the fact that we already know a! We can plug ~ -0.14 in, then solve with substitution. (I hope you've done systems of linear equations)

I got -0.14x^2  - 0.04x + 3.58, which has a max at (-0.134, 3.583) but the second x-int had shifted by about 0.3 ft, probably thanks to only using 2 decimal places. I put a ~ for "approximately equal to" on every number in this post that is approximated to 2 decimal places.

EDIT: I may also have messed up copying over some numbers.

Edited by Mad Rocket Scientist
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Posted (edited)

A physics ecyclopaedia is obviously required.

If the ballista moves back significantly, I would try to put a thick physics book behind the wheeled ballista to realize the exact point at which it throws the projectile.
(Yes, several more attempts).
Say, if the ballista stops at 1 m behind its original position, I would try to put the book at 50 cm, 25 cm, 40 cm, 37 cm, ... to realize that the projectile leaves the launcher at 36 cm behind the original ballista position.
I mean, when the flight distance stops changing, this is obviously the point where the ballista and the stone are separated.

Another set of shots I would do with a, say, sheet of cartoon (say, the same physics book) placed somewhere above the highest point of trajectory, to realize at which altitude the stone stops hitting the cartoon.

This could make calculation more accurate as you know the distance and altitude more precisely.

The same altitude test with the non-wheeled ballista, as the real trajectory can differ from the ideal parabola.

Another set of shots with stones of different mass to realize how much the impulse/kinetic energy changes, because of energy wastes.
Measuring also every time the ballista recoil distance.

Probably another set of shots with the heaviest projectile and additional weights on the ballista to make a table and a diagram how does the mass of ballista effect the recoil distance (and so, the ballista velocity).

Ideal equations are highly inaccurate here, and unlikely can be used directly. They can mostly just describe the whole plot of the experiment and a type of empiric formula you get at the end.

Edited by kerbiloid

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