Free Falling Objects: Find Time for Mailbag to Reach Ground

  • Thread starter Thread starter pberardi
  • Start date Start date
  • Tags Tags
    Falling
AI Thread Summary
The height of a helicopter is modeled by the equation h = 3.00t^2, and after 2 seconds, it releases a mailbag. The initial height at release is calculated to be 12 meters, and the initial velocity is determined to be 24 m/s. Upon correcting the height function to h = 3t^3, the initial velocity becomes 36 m/s. The equation for the mailbag's descent is set up as 0 = 24 + 36t + 0.5(-9.8)t^2, leading to a solution of approximately 7.96 seconds for the mailbag to reach the ground. Proper unit notation is emphasized for clarity in calculations.
pberardi
Messages
16
Reaction score
0

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

Homework Equations


Xf = Xi + VxT + .5at^2


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.
 
Physics news on Phys.org
pberardi said:

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

Homework Equations


Xf = Xi + VxT + .5at^2


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.
 
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?
 
pberardi said:
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.
 
Should be
0 = 24 + 36t + .5(-9.8)t^2
t = 7.96
?
 
pberardi said:
Should be
0 = 24 + 36t + .5(-9.8)t^2
t = 7.96
?
That looks good.
 
Thank you sir.
 
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:
Back
Top