# Homework Help: Vertical Motion of Objects

1. Oct 6, 2009

### ryno2107

The height of a helicopter above the ground is given by h = 3.50t^3, where h is in meters and t is in seconds. After 1.85 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? (Ans. in s.)

(1) y-y0=v0t+1/2*gt^2

(2) v^2=v0^2+2g(y-y0)

(3) y-y0=1/2*(v0+v)t

I sub. 1.85s for t in h=3.50t^3 and h=y0=22.2m.. This is the initial position. The final position is y=0. So I substituted all of my known values into (1) and got v0= -2.92m/s(initial velocity)
then I sub. all the known values into (2) and got v= -21.1m/s. Finally, I sub. v, v0, and the rest of the known values into (3) and got 1.85s. I believe it is the wrong answer. Please help me to solve this problem.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 6, 2009

### Delphi51

I agree with your initial height.
But you can't use a constant acceleration formula to find the velocity of the helicopter at time 1.85 s. There are no formulas for height increasing as the cube of time. If you know a bit of calculus, you can differentiate to get a formula for velocity. If not, you'll have to carefully graph y = 3.50t^3 and draw a tangent line at time 1.85 to estimate the slope, which is the velocity.