When Does the Package Reach the Ground?

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SUMMARY

The discussion centers on calculating the time it takes for a package to reach the ground after being dropped from a helicopter ascending at 5.5 m/s from a height of 100 meters. The relevant physics equation used is delta y = v0t + 1/2at², where v0 is the initial velocity (5.5 m/s), a is the acceleration due to gravity (-9.8 m/s²), and delta y is the change in height (-100 m). The correct solution involves using the quadratic formula to solve for time, yielding an answer of approximately 5.11 seconds, as confirmed by the teacher.

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Homework Statement


A helicopter is ascending vertically with a speed of 5.5m/s. At a height of 100m above the Earth, a package is dropped from a window. How much time does it take the package to reach the ground?


Homework Equations


delta y = volt + 1/2at^2


The Attempt at a Solution


well I know that:
Vo= 5.5m/s
a=-9.8 m/s^2
delta y= -100m

I re-wrote the equation so that I would solve it using the the quadratic formula.
-5.5 + the square root of (5.5)^2 -4(-4.9)(100)/ 2(-4.9)
My teacher gave us the answer of t= 5.11s, but I can't figure out how he did it.
 
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Here's a site with the full quadratic formula: http://mathworld.wolfram.com/QuadraticFormula.html" . Just plug in the numbers and you'll get 2 solutions. Only one will make sense for this problem.
 
Last edited by a moderator:
Your equation is not quite correct it's -5.5 plus or minus the sq rt...etc. You will get 2 values for t...which is correct? (also, be sure to check all math).
 

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