# One Dimensional Kinematics

• RJLiberator

Gold Member

## 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.

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:
• RJLiberator
Thanks for the help, that makes sense indeed.

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