Question

The height of a helicopter above the ground is given by h = 2.80t^3, where h is in meters and t is in seconds. At t = 2.50 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer 1

First, you need to calculate the height when it gets dropped. Can you do that part?

Answer 2

yeah, height should be 43.75

Answer 3

Right. So from there you can just use one of the kinematic equations:\[d=V_ot+\frac{1}{2}gt^2\]Here, V0 is 0 m/s since it's being dropped. So you end up with this equation:\[43.75m=\frac{1}{2}(9.8m/s^2)t^2\]Now you just need to solve it for t.

Answer 4

t should be t=2.988, but webassign says its wrong... Apparently the correct answer is 11.5

Answer 5

With the information given, there's no way it would 11.5 seconds. That doesn't even make logical sense.

Answer 6

then ask webassign, cause it's saying 2.988 is wrong

Answer 7

Maybe there's a typo in the question...either by you or in webassign. That's all I can think of. This is a very basic physics question.

Answer 8

i thought so too until i couldnt even get the "right answer"

Answer 9

Wait...I think this is supposed to be a projectile motion question. That means you also have to account for the upward velocity of the helicopter in that it reached a height of 43.75m in 2.0s.

Answer 10

sadly, i still dont know what to do from there

Answer 11

Gimme a minute...I'll do it that way. I made the incorrect assumption that the helicopter was hovering at the point at which the mailbag was dropped.

Answer 12

\[H_0=2.80t^3=43.75m\]\[t=2.50s\]\[V_{0} = 8.4t^2=8.4(2.80)^2=52.5m/s\]The max height will be :
\[H_{max}=H_0+V_0t+\frac{1}{2}gt^2\]\[0 = 43.75m + 52.5m/s + \frac{1}{2}(-9.8m/s^2)t^2\]Solving for t you end up with the times where it's at max height as well as when it hits the ground.

Answer 13

You still there?

Answer 14

yeah..

Answer 15

Sorry...I had the correct value the first time. I'm just tired...
\[0 = 43.75m + 52.5m/s(t) + \frac{1}{2}(-9.8m/s^2)t^2\]\[t=11.5s\]

Answer 16

Ok...now that I'm done screwing up...do you understand what I did?

Answer 17

wish i did.. haha

Answer 18

i see that you calculated the velocity when the bag hit the ground using the derivative, etc
but i dont see how you pulled 11.5 out of that equation

Answer 19

Ok, we know the starting height is 43.75m.

Since the helicopter has an upward velocity when the mailbag is dropped you need to calculate what that is. Take the derivative of 2.80t^3 to get 8.4t^2. Plugging in 2.5s as t gets you 52.5m/s.

The rest is just using the kinematic equation:\[\large h_{final}=h_{initial}+V_0t+\frac{1}{2}gt^2\]

Answer 20

using that equation is still get 5.976

Answer 21

http://www.wolframalpha.com/input/?i=solve+for+t%3A+0+%3D+43.75+%2B+52.5t+%2B+ \frac{1}{2}%28-9.8%29t^2

Answer 22

Copy the entire link...not sure why it didn't paste correctly.

Answer 23

okay ill work on it, thanks

Answer 24

At this point you just need to solve for t correctly using my last equation. You should get 11.491 which is about 11.5s.

Answer 25

np