- #1
Vii
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Homework Statement
The height of a helicopter above the ground is given by [tex]h=3.00t^3[/tex], where h is in meters and t is in seconds. After 2.00s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
Homework Equations
[tex]d=v_1t + \frac{1}{2}at^2[/tex]
The Attempt at a Solution
First, I figured out what the height would be at 2.00s by substituting t with 2.00s.
[tex]h=3.00(2.00)^3[/tex]
[tex]h=24m[/tex]
Then I took [tex]d=v_1t + \frac{1}{2}at^2[/tex] and plugged in 24m at d and assumed [tex]v_1[/tex] as 0.
[tex]24=0t + \frac{1}{2}at^2[/tex]
[tex]24=\frac{1}{2}at^2[/tex]
And plugged in 9.8m/s^2 into a.
[tex]24=\frac{1}{2}9.8t^2[/tex]
Which gave me t=2.21s
The answer is wrong and I cannot figure out how to do this problem.
I think my assumption of [tex]v_1=0[/tex] is wrong. Furthermore, I think my assumption that the helicopter stops ascending at the time the package is dropped is also wrong. I've racked my brain on this question for a good 2 days now and I still can't figure out how to arrive at the answer.