Please Help Me With This Physics Exercise

[mufarrid]

A chopper drops boxes of relief goods to a group of people stranded on an island. The goods were released 20.0 m above the sea and landed 40.0 m from a point exactly below where the goods were released. What was the velocity of the goods when they were released? What was the velocity of the chopper?

[jim360]

Jim isn't a lecturer or teacher currently, but they'll get around to this problem shortly anyway :)

Thanks for the shout-out, I might have missed this otherwise!

[Buenchico]

>>> Jim isn't a lecturer or teacher currently, but they'll get around to this problem shortly anyway

Hopefully with the same result that I posted at 1655 ;-)

[jim360]

Probably, although imo you've played very fast and loose with minus signs :P

Also, whilst I suppose "released" is meant to clue you in to the idea that the initial vertical speed is zero, I'm not a fan of just assuming that without stating it.

5/7 for the right answer though, just gotta work on the presentation :P

[Peter Pedant]

chris - how does it feel to give the right answer + diag

and the usual suspects continue with the usual fog " foo we need Jim we do, where Jim den?"

[Peter Pedant]

Jim is my favourite quantum theorist

[jim360]

So, anyway. Hope this is clear. I take my time, but hopefully the advantage of discussing is that the thought process becomes a bit clearer. If not, feel free to come back with questions (or just copy Buen's answer :P )

in the rest of this post, g = 10, because I can't be bothered with numberwork.

To set the problem up, firstly, we only have the suvat equations: as Buenchico says, let's use the one "s = ut + 1/2 a t^2".

Question 1: why did Buen (and I) know to pick this one?
Answer 1: Because we are asked about the *initial* velocity (u), are given distances (s) and the acceleration (a), and aren't interested in the *final* velocity (v). So this clues us in to using that particular equation.

suvat luckily works in both the horizontal (x) and vertical (y) directions separately! So we can just write down two copies of this equation for the given information.

Now, let's make a choice. We need to choose: where is the starting point of the motion? And where, relative to this, is the end point?

I choose the helicopter to be the start point (0,0), and up to be positive. So this means that the goods *fall* 20 metres. That's a negative change in height! down = negative. Also, gravity pulls down so, again, gravity is negative here!

It's key to get these signs right. So the y equation is:

s = -20, u = ?? (call it uY), t = ??, a = -10, sub in:

-20 = uY t - (1/2)*10*t^2 (Equation 1)

There's too much we don't know here, but before moving on, we should ask what is uY really?

Question 2: What does it mean to "release" the goods?
Answer 2: It means that the helicopter just "dropped" them. It didn't push, or throw, or anything. This means that there's no initial downwards speed, so we can just say uY = 0!

Now let's think about the x motion. Now there's no acceleration, remember! gravity only acts down, not left or right. Also, the goods land 40 metres away, so we have this time:

s = 40, u = ?? (call it uX), t = ??, a = 0:

40 = uX t (Equation 2)

Question 2: Why did I use the same symbol for t both times?

Answer 2: Because when it hits the ground, the same amount of time has passed to cover 20 metres down and 40 metres along!

So we can put the two equations together:

-20 = - (1/2)*10*t^2 = - 5 t^2 (Equation 1)
40 = uX t (Equation 2)

As luck would have it, those annoying minus signs cancel, and we can also divide both sides of equation 1 by 5, so it's really just:

4 = t^2
40 = uX t

At this point, Buen's answer differs from mine only because he used a different (better) value of gravity, but the sneaky advantage of my choice is that Equation one is now only asking "what is the square root of 4?", which is just 2. So t = 2, and we put that into equation 2, giving:

t=2
40 = uX*2
uX = 40/2 = 20.

So the initial velocity was 20 m/s in the horizontal direction only.

* * * * * *

And what about the chopper? Again, if the goods were "released", then the helicopter didn't do anything to the goods other than "let go". So it must have been moving in the same way, ie the goods and the chopper were both moving 20m/s horizontally when the copter dropped them.

((We could also have guessed this by using the technique of "proof by I wouldn't have a clue otherwise":

1. Suppose the speed of the helicopter had nothing to do with the speed the goods were moving at.
2. Then there's no way to answer the problem.
3. So they must be the same.

NB do not use this method in a test :P))

[Buenchico]

I've been waiting for Jim to come along and complicate things ;-)

[jim360]

Hah!

But even at A level, being careful with your choice of coordinates and directions is worth stressing. Choosing down = positive is of course fine but you should probably at least have said that's what you were doing :)

[jourdain2]

Glad to be of help, Jim. :)