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Dropping a package with initial velocity

  • Thread starter Calpalned
  • Start date
  • #1
297
6

Homework Statement


A helicopter is ascending vertically with a speed of 5.30m/s . At a height of 100m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: v0 for the package equals the speed of the helicopter.]

Homework Equations


1) vf2 = v02 + 2aΔx
2) vf = v0 + at
3) xf = x0 + v0t + ½(a)t2
4) Quadratic Formula

The Attempt at a Solution


Using the third equation with Δx = xf - x0 = -100, v0 = 5.3 m/s and acceleration = -9.81, and the quadratic formula, I get the right answer of t = 5.09 seconds. However, I don't get why this other logical method does not work. Using equation one, I substitute v0 with 5.3, a with -9.81 and Δx with -100. This leads to vf2 = 5.32 + 2(-9.8)(-100). vf = -44.598 m/s. Plugging this value into the second equation I get -44.58 = 5.3 - 9.8t. t equals 4.09 seconds. What did I do wrong?


 

Answers and Replies

  • #2
Nathanael
Homework Helper
1,650
239
[itex]\frac{44.58+5.3}{9.8}=5.09[/itex] seconds
 
  • #3
297
6
Thank you so much! I see: I subtracted incorrectly
 

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