Finding position and Velocity of a dropped Sandbag from 40.0m above the ground

[Toranc3]

Homework Statement

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.

Homework Equations

V=Vo + A*t

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

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

Thanks in advance!

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

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?)

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.

 

[clamtrox]

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

 

[Toranc3]

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

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!

 

[Toranc3]

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.

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!