# One Dimensional Kinematics

1. Jan 19, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
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.

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

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.

2. Jan 19, 2015

### Nathanael

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:
Yes, that was also correct.

Last edited: Jan 19, 2015
3. Jan 19, 2015

### RJLiberator

Thanks for the help, that makes sense indeed.

4. Feb 2, 2015

### RJLiberator

I wanted to bump this post for any more opinions before I complete this assignment.