One Dimensional Kinematics

  • #1
RJLiberator
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Homework Statement


As a helicopter carrying a crate takes off, its vertical position (as well as the crate's) is given as: y(t)= At^3, where A is a constant and t is time with t=0 corresponding to when it leaves the ground. When the helicopter reaches a height of h = 15.0m the crate is released from the underside of the aircraft. From the time the crate leaves the helicopter to the time it hits the ground, 2.50 seconds pass. Calculate A.

Homework Equations


Kinematic equation: S = Si + Vi*t+1/2*a*t^2
given equation: y(t)=At^3

The Attempt at a Solution


I first found the initial velocity that the crate left the aircraft using kinematics:
0=15+Vi*2.5+1/2*-9.81*2.5^2
Vi = 6.2625 m/s

Next, I took the derivative of the position function and got the velocity function y'(t)=3At^2

I then set 6.2625 = 3A*t^2
and noticed that using the first equation A=15/t^3
6.2625=3*15/t^3*t^2
simplified to
6.2625=45/t
t=7.1865

I then inputted this t time back into the original equation to find A
15=A*7.1865^3
A=0.0404

I added units of m/s^3 to the answer so that it cleared out in units.

Does this seem right?
Is it okay to add the units at the end here as m/s^3?

I don't like the answer A=0.0404, it seems to small.
 

Answers and Replies

  • #2
Nathanael
Homework Helper
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It all looks good to me. It's okay that it turned out small, it would be worse if it turned out big; say A=1, then in just 10 seconds the helicopter would be moving vertically at 300 m/s!

Edit:
I added units of m/s^3 to the answer so that it cleared out in units.
...
Is it okay to add the units at the end here as m/s^3?
Yes, that was also correct.
 
Last edited:
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  • #3
RJLiberator
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Thanks for the help, that makes sense indeed.
 
  • #4
RJLiberator
Gold Member
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I wanted to bump this post for any more opinions before I complete this assignment.

I feel "good" about my answer, but I don't feel conceptually happy about it.
 

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