# Homework Help: Physics Falling Bodies Question

1. Sep 13, 2007

### Phan

1. The problem statement, all variables and given/known data
A person in a hot air balloon that is rising accidentally drops a can of paint from the balloon. The balloon is moving up at a constant velocity of 4m/s, the distance to the ground is at the moment 4m. Find the time it takes for the can to drop to the ground.

2. Relevant equations
d=v1*t+(1/2)at^2
v2^2=v1^2 + 2ad
a = (v2-v1)/t

3. The attempt at a solution
To try to solve the equation, I first started off by using the first formula and rearranging to solve for (t) while leaving v1 as 0, and using 9.8m/s (grav) for (a). However, I quickly realized that v1 is actually 4m/s, so I quickly changed my solution.

After I solved for (t) again with the new value for v1, I got a really small number. This seemed wrong, so I tried instead the second equation to solve for v2, then I input that into equation 3, but it was also a small number, I got 0.58. My friend got 1.4s instead, and I am pretty sure he is right.

Can anyone point me in the right direction? I didn't review physics 11 yet, and I am struggling to recall the procedure for the initial motion questions. Thanks for any help.

2. Sep 13, 2007

### learningphysics

Yes, use that first formula... careful about signs and directions...

3. Sep 13, 2007

### Phan

When I rearranged the formula to solve for (t), I got:

2d/(v1*a) = t^3

Assuming that I the (ts) added up to an exponent of three, I took the root of the left side and was left with an odd number.As I plugged in everything (with negative acceleration), I got a time of 0.58s. Am I rearranging the formula the wrong way then?

4. Sep 13, 2007

### learningphysics

YEah, that isn't right... careful with the algebra... can you show all your steps... I can poit where it's going wrong.

5. Sep 13, 2007

### Phan

Ok, I thought I was doing something wrong but I wasn't sure what.

d = v1t + (1/2)at^2
2d/v1 = t + at^2
2d/v1(a) = t + t^2

Sorry if my algebra is a little rusty, I haven't touched it in a long time. I figure that it is either step 2 or 3 where I went wrong, but I'm not sure of any other way of doing it.

6. Sep 13, 2007

### learningphysics

Yeah, the second step is wrong... multiply both sides by 2... do that as one step... then if you divide by v1... you have to divide everything by v1 (all 3 terms)...

you don't really have to rearrage the equation... just plug in the numbers and solve for t... it's a quadratic equation.

7. Sep 13, 2007

### Phan

I'm still not sure what you mean by multiplying both sides by 2 and doing it as one step. Would the next step be:

2d = 2(v1t + 1/2at^2)
2d = 2v1t + at^2
2(4) = 2(4)t + (-9.8)t^2

I'm sorry, but I have forgotten quite a bit about algebra and can't remember how to specifically rearrange.

8. Sep 13, 2007

### learningphysics

Yes, that's right. And if you want to divide by v1... you'd have to divide all 3 terms by v1...

Do you remember the quadratic formula? Simplify the above equation...

9. Sep 13, 2007

### Phan

Ok, I just wanted to make sure I was doing it correctly first. I had forgotten that pulling a single term (from 1/2a) wouldn't really work out too well.

2d = 2v1t + at^2
8 = 8t + (-9.8)t^2
(-9.8t^2) + 8t - 8 = 0

However, if I plug that into the Quadratic Formula, I would get a negative under the root sign?

10. Sep 13, 2007

### learningphysics

Ah yes... good catch... the d should be -4 (we're taking up positive, down negative)... because it's a displacement downwards of 4m...

11. Sep 13, 2007

### Phan

Hrmm, I thought the displacement wouldn't matter, but clearly I have to give everything a direction from a reference point...

From here, I would plug it into the Quadratic Formula:
(-9.8t^2) + 8t + 8 = 0
b^2 - 4ac
8^2 - (4)(-9.8)(8)
ROOT-> 377.6
->19.43

The equation is then:

X= [(-b) +- (19.43)]/2(-9.8)
= [(-8) +- (19.43)]/(-19.6)
= 1.4s

Ahhh woooww.... Thanks so much for your help, I wouldn't have been able to do it if not for your guidance. I can't believe I forgot so much about basic algebra, I suppose it's time for a review.

Once again, thanks a lot for your help. :)

12. Sep 13, 2007

### learningphysics

No prob... with a little practice your algebra will back up to speed easily.

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