Please Help

[CanadianChick 0]

Since there isn't an Education Folder yet, I thought I'd put it here. I'm real stuck on a question, and it seems so simple, but I can't get it. So all of you guys who are good, or even decent, at physics, help a girl out.

 

A projectile is fired horizontally with an intial velocity of 13.0m/s at an object 256m away. If the object starts to fall at the same instant the projectile is fired, how far will the object fall before it is struck by the projectile?

 

For the love of god, please help.

[CanadianChris 0]

Sure. I may be a little rusty, but here goes...

 

In essence, the question we are trying to answer is, how far will the projectile have fallen once it has travelled 256m? (Because once it has travelled that far, we assume it will meet the object, which is falling perfectly vertically. This is a critical assumption -- without it, we can't solve the problem.)

 

The formula for displacement of an object in one dimension is:

 

d = v(i) * t + 1/2 * a * t^2

 

where:

d = displacement

v(i) = initial velocity

a = acceleration

t = time

 

In two dimensions, though, there are two formulas, one each for the horizontal and vertical components:

 

x = v(ix) * t + 1/2 * a(x) * t^2

y = v(iy) * t + 1/2 * a(y) * t^2

 

where:

x = horizontal displacement

y = vertical displacement

v(ix) = initial horizontal velocity

v(iy) = initial vertical velocity

a(x) = horizontal acceleration

a(y) = vertical acceleration

t = time.

 

OK, now we know these things:

 

v(ix) = 13 m/s (the speed the projectile was fired)

v(iy) = 0 m/s (when fired, the object travels perfectly horizontally)

a(x) = 0 m/s^2 (assuming air resistance is negligible, there is no acceleration or deceleration of the projectile once it is fired)

a(y) = -10 m/s^2 (force of gravity on the projectile)

x = 256m (the distance between the projectile and the object)

 

We are trying to find y (how far the projectile -- and thus the object -- have fallen). We do not know t, but we can calculate it from the first formula:

 

x = v(ix) * t + 1/2 * a(x) * t^2

256 = 13 * t + 1/2 * 0 * t^2

256 = 13 * t

t = 256/13 = 19.7 s.

 

Now we put that value into the second equation:

 

y = v(iy) * t + 1/2 * a(y) * t^2

y = (0 * 19.7) + (1/2 * -10 * 19.7^2)

y = 1/2 * -10 * 388.1

y = -1940.5 m.

 

So, the object falls 1940.5 m before being hit by the projectile.

Edited December 14, 2003 by CanadianChris

[Guest cobainwasmurdered]

what he said.

[CanadianChick 0]

Thanks man. You should be my teacher instead of the one I have now.

[CanadianChick 0]

Oh, god damn. I looked at the back of the book and the answer is 1200. Ugh, I should have dropped this course when I could of.

[CanadianChris 0]

1200????? What the hell?

[{''({o..o})''} 0]

98.5 is pretty close to 1200...

[CanadianChick 0]

Whatever. I just have to pass this year. It's not like I'm going into physics 12. It's weird; I'm so good at math, but am absolutely brutal at physics. Go figure.

[CanadianChris 0]

Found my original mistake...forgot to square time in the second equation. Still doesn't come up with the right answer, though...hmmm.

[CanadianChris 0]

Whatever. I just have to pass this year. It's not like I'm going into physics 12. It's weird; I'm so good at math, but am absolutely brutal at physics. Go figure.

I hear you. I was never really brutal at physics, but I absolutely hated it. That's why I went into statistics in university...nothing even remotely resembling physics there.

 

Oh, and I ran 1200 through those equations, and there's no way that can be the answer if the numbers you gave in the initial problem are correct. Absolutely no way.

[CanadianChick 0]

Maybe it was just a mistake in the book than. Ah well. I'll do alright. If I get a B on this test, I'll be happy.

[Slayer 0]

Chris' explanation just seems a bit too complex for a basic HS physics class (I sure as hell don't remember doing something that "complex")

 

Unless CC is taking a more advanced course

[CanadianChick 0]

I'm not taking an advanced class, but I understood his explaination. I find that better than when my teacher tries to dumb it down, but just ends up screwing everybody up.

[CanadianChris 0]

I was surprised they were doing advanced projectiles in a high school physics class, to be honest.

 

Oh well. Good luck CC!

[Slayer 0]

Maybe it's a Canadian thing

[Jingus 0]

The question is impossible to realistically answer, since they didn't specify what kind of "object" it was. A bowling ball and a feather fall and two completely different rates of speed.

[CanadianChick 0]

The question is impossible to realistically answer, since they didn't specify what kind of "object" it was. A bowling ball and a feather fall and two completely different rates of speed.

I think we're suppose to assume that has no effect on the question.

[CanadianChris 0]

The question is impossible to realistically answer, since they didn't specify what kind of "object" it was.  A bowling ball and a feather fall and two completely different rates of speed.

I think we're suppose to assume that has no effect on the question.

And it doesn't. Two objects dropped from the same height will accelerate through the air at the same speed - 10 m/s/s.

[Guest cobainwasmurdered]

Maybe it's a Canadian thing

I'm canadian and I have no clue what they're talking about.

[NYU 0]

The question is impossible to realistically answer, since they didn't specify what kind of "object" it was. A bowling ball and a feather fall and two completely different rates of speed.

Well, in the problem, you have to assume that it's falling in a vacuum, where everything falls at the same speed, with an acceleration of 9.81 m/s^2. In a vacuum, a bowling ball and a feather would reach the ground at the exact same time. If it's just thrown off a roof, obviously the wind would make the feather fall after the bowling ball. But for physics problems, you just have to assume there is absolutely no wind, which is pretty much a vacuum.

 

Ugh, I just wasted Post 900 on a Physics problem....and I sucked at Physics. Fucking 68 on the NYS Regents....

[Your Paragon of Virtue 0]

The question is impossible to realistically answer, since they didn't specify what kind of "object" it was.  A bowling ball and a feather fall and two completely different rates of speed.

I think we're suppose to assume that has no effect on the question.

And it doesn't. Two objects dropped from the same height will accelerate through the air at the same speed - 10 m/s/s.

You're both sort of correct. The velocity of the ball will be greater, due to a variety of factors, including it's mass, air resistance, etc. Acceleration may not be 9.8 m/s/s. I believe for this question you should assume that air resistance is negligible however.

 

We never did this stuff in Physics 11 either, but now I'm in Physics 12 and they just dumped all this proj. motion shit on us. It's okay though, it and the rest of the curriculum is fairly simple, since it involves no memory or hard work whatsoever. As for what she asked, it is actually considered a basic question, as it states that the projectile is fired horizontally, as opposed to being fired at an angle.

 

Odds are the back of the book is wrong, as CC seemed to use the equations of motion pretty well. And by CC, I mean CanadianChris.