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Old Jun 25th 2014, 03:37 PM   #1
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Helicopter dropping a package

I am having trouble determining the signs (+ or - ) for acceleration and velocity. Here is the question:

A helicopter is ascending vertically with a speed of 5.10 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: v(initial) for the package equals the speed of the helicopter.]

I know I have to use the following equation to solve the problem: y = y(initial) + v(initial)t + 1/2(a)(t^2).

I have determined that y(initial) is +105 m, y = 0. Because the package is falling downwards, I will make the velocity -5.10 m/s. Because the package is picking up speed right before it hits the ground, the sign of acceleration should be the same as the sign of velocity. Thus, a = -9.8 m/s/s. However, when I plug my answers into the equation, I do not get the right answer. Can someone explain to me what is wrong with my logic?
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Old Jun 26th 2014, 06:06 AM   #2
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You need to be consistent with the sign of y. I recommend always picking positive y as increasing altitude, and hence -y is decreasing altitude. So yes, y_initial = +105 and y_final = 0. With this convention positive velocity means the object is rising, and negative velocity means it's falling. The initial velocity of the box is upward at 5.1 m/s, so it's value is positive, not negative. Finally, acceleration is positive if upward velocity is becoming greater (or negative velocity is decreasing in magnitude), and acceleration is negative if the object's negative velocity is increasing in magnitude, or it's positive velocity is decreasing. Thus the value for g is negative. Putting it all together, you have:

y_final = y_initial + v_it + (1/2)at^2:

0m = 105m + (5.1 m/s)t - (1/2)(9.8 m/s^2)t^2

Hope this helps.
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Old Jun 26th 2014, 09:34 PM   #3
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Thank you!
I appreciate the very straightforward you outlined everything. It makes sense now!
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