# Homework Help: Finding position and Velocity of a dropped Sandbag from 40.0m above the ground

1. Aug 2, 2012

### Toranc3

1. The problem statement, all variables and given/known data

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00m/s , releases a sandbag at an instant when the balloon is a height v = 40.0m above the ground . After it is released, the sandbag is in free fall.

A: Compute the position and velocity of the sandbag at .250s and 1.00s after its release.

2. Relevant equations

V=Vo + A*t

Y=Yo + Voy*t + (1/2)Ay*t^(2)

3. The attempt at a solution

Sorry, picture is up now
http://www.freeimagehosting.net/t/v8sxq.gif

A: At 0.250s

V=Vo + A*t

V=5.00m/s -(9.8m/s^(2) *(0.250s)) = -2.55m/s or 2.55m/s upward...[Answer is actually +2.55m/s but the velocity is going downwards so it is negative. (is that correct?)

Y=Yo + Voy*t + (1/2)Ay*t^(2)

Y=40.0m + 5.00m/s(0.250s) + 1/2*(-9.8m/s^(2))*(0.250s)^(2)=40.9m

At 1.00s

V=Vo + A*t

V=5.00m/s -9.8m/s^(2)(1.00s)=-4.8m/s

Y=Yo + Voy*t + (1/2)Ay*t^(2)

Y=40.0m +5.00m/s(1.00s) +1/2*(-9.8m/s^(2))(1.00s)^(2)=40.1m

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Aug 2, 2012
2. Aug 2, 2012

### Simon Bridge

When you do these problems, it is good to draw a picture. On your picture you draw an arrow indicating the positive direction. That way you won't lose track.

Looking at your calculation, you put v0=+5m/s ... so the positive direction is "upwards".

In a quarter of a second, the sandbag is still moving upwards ... so its velocity should be positive. You substituted the values into the equation right but something got switched in the arithmetic.

3. Aug 2, 2012

### clamtrox

That looks fine. Did you have anything in particular you were wondering about?

4. Aug 2, 2012

### Toranc3

Well I was not sure whether the velocity at .250s, which is 2.55m/s should be positive or negative. I made my upward positive. I believe Simon Bridge answered it. At a quarter of a second the velocity is still moving upwards. Thanks!

5. Aug 2, 2012

### Toranc3

Oh I see. Yeah I made my positive upwards and I did draw a picture. I just was not sure if the velocity at .250 seconds should be + or - 2.55m/s. I understand what you said though. Hey thanks!