# Homework Help: Acceleration Time Help

1. Sep 27, 2015

### njurcik

1. The problem statement, all variables and given/known data
A helicopter rises vertically with a constant upward acceleration of 0.34 m/s2 . As it passes an altitude of 20 m, a wrench slips out the door.

2. Relevant equations
How soon does the wrench hit the ground?
At what speed does the wrench hit the ground?

3. The attempt at a solution

I have attempted to solve for the initial velocity of the wrench (3.6m/s), and from that I was able to solve for time using x=x(0) +V(0)t+1/2at^2
From this I get that the time is then 2.4s.

I am a bit confused tho, as first I know this is wrong. And second whether we are calculating the time of the item travelling up and then back to the ground again.

For the second part of the question, I know I need to find the final velocity of the wrench on its way down
for this i would use v2^2=v1^2+2AD
So for this one I have been getting 19.8m/s. But again I am not sure if this is the correct answer.

If someone can please lend a hand on this that would be perfect.

2. Sep 27, 2015

### njurcik

Sorry the relevant equations would be in the third section of the post - sorry about that

3. Sep 27, 2015

### Staff: Mentor

You should get in the habit of keeping a few more decimal places for values that you will be doing further calculations with. These so-called "guard digits" will help prevent rounding and truncation errors from creeping into the significant figures of future calculations.

That said, your initial velocity and time look good.
It's not clear to me what values you are using in your formula v2^2=v1^2+2AD, so it's hard to comment on your method. If you are using the right values you should get a correct result (your result is close to but not exactly what I would expect). However, you can use another kinematic formula to check your results: you know the initial velocity, the acceleration, and the time of flight, so what formula might you use?

4. Sep 27, 2015

### Mister T

How do you know it's wrong?
You're calculating the time it takes for the wrench to undergo a displacement of -20 m. Note that the distance the wrench travels is greater than 20 m because it travels up and then back down again. However, the distance isn't relevant. It's only the displacement that's relevant.

I agree that you have to be more careful about round-off errors. Other than that you're doing well.