Sandbag is dropped from moving ballon

[rsala]

nevermind, I've solved it, thanks anyway

hello, I've tried this simple problem a few times and mastering physics keeps taking points off =(.
ive never used latex so bare with me

Homework Statement

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground View Figure . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.

Compute the velocity of the sandbag at a time 1.15 s after its release.

Homework Equations

x = X$$_{0}$$ +V$$_{0}$$T + .5aT$$^{2}$$
V$$^{2}$$ = V$$^{2}_{0}$$ + 2a(x - x$$_{0}$$)

The Attempt at a Solution

I assume that the initial velocity of the sandbag once it is released is 5ms since that was the velocity of the balloon.

first i found the position of the sandbag at 1.15.
x = 40 + 5(1.15) + .5(-9.81)(1.15)$$^{2}$$
x = 39.3

up to this point i am 100% sure i am correct.

next i inserted this information into the velocity formula.
V$$^{2}$$ = V$$^{2}_{0}$$ + 2a(x - x$$_{0}$$)

V$$^{2}$$ = (5)$$^{2}$$ + 2(-9.81)(39.3 - 40)
$$\sqrt{V^{2}}$$ = $$\sqrt{38.688}$$
V = 6.22

Mastering physics says incorrect and to check my signs,, I've checked my signs i can't find my error

edit: ops, since its falling i take the negative root not the positive root, problem solved..answer was -6.27

 

[anathema]

I'm glad to hear that you were able to solve the problem and find the correct answer! It's always a great feeling when you are able to figure out a difficult problem. In case you were still curious about where you went wrong, it looks like you may have forgotten to take into account the initial velocity of the sandbag when solving for its position at 1.15 seconds. Since the sandbag is released from a moving balloon, it already has an initial velocity of 5 m/s in the upward direction. This would change your equation for position to:

x = 40 + 5(1.15) + 0.5(-9.81)(1.15)^2 = 39.3 m

Then, when solving for the velocity using the equation V^2 = V_0^2 + 2a(x-x_0), you would get:

V^2 = 5^2 + 2(-9.81)(39.3-40) = 38.688 m/s^2

Taking the square root of this value would give you the correct answer of -6.27 m/s, which is the same as what you found when you realized your mistake.

I hope this helps and keep up the good work in your studies!

 

[Moetasim]

Hello,

I'm glad to hear that you were able to solve the problem and find your error. It's important to always double check your signs when working with equations and make sure they align with the direction you have chosen as positive. Keep up the good work!