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Free falling objects

  • Thread starter pberardi
  • Start date
16
0
1. Homework Statement

The height of a helicopter above the ground is given by h = 3.00t^2, 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

2. Homework Equations
Xf = Xi + VxT + .5at^2


3. The Attempt at a Solution
h = 24
a = -9.8
Vi = 6.00t^2 at 2 = 24 m/s not sure about this

0 = 24 + 24t + .5(-9.8)t^2
t = 5.7

Did I go about this right? Please help.
 
109
0
1. Homework Statement

The height of a helicopter above the ground is given by h = 3.00t^2, 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

2. Homework Equations
Xf = Xi + VxT + .5at^2


3. The Attempt at a Solution
h = 24
a = -9.8
Vi = 6.00t^2 at 2 = 24 m/s not sure about this

0 = 24 + 24t + .5(-9.8)t^2
t = 5.7

Did I go about this right? Please help.
Wouldn't the initial height be 12? 2^2*3 = 12?

Also, do you do physics with calculus? If so, you could take the derivative of the height function to get the velocity function, and simply plug in for t. If not, you can only get the average velocity, which won't do you any good in this problem, since the helicopter is accelerating.
 
16
0
I am sorry, my mistake. I did do the calculus. But I wrote the height function incorrectly. It is supposed to be h = 3t^3

So based on this new information, is it correct?
 
109
0
I am sorry, my mistake. I did do the calculus. But I wrote the height function incorrectly. It is supposed to be h = 3t^3

So based on this new information, is it correct?
The initial height is correct, but the derivative of 3t^3 is 9t^2. With that in mind, the initial velocity would be 36.
 
16
0
Should be
0 = 24 + 36t + .5(-9.8)t^2
t = 7.96
?
 
16
0
Thank you sir.
 
536
2
They might deduct a couple points if units aren't shown in every step ...
0 = 24.0 m + (36.0 m/s) t + (1/2) (-9.8 m/s^2) t^2
 
Last edited:

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