Mechanics difficulties - acceleration

1. Dec 15, 2003

Micah18

Hey, I came across this site on one of my many google searchs on my quest to solve this problem, and was hoping someone could help me out.

The following word problem is the one I am trying to solve:

An object is dropped from 100 feet up. Negating wind resistance, calculate the time it took for the object to hit the ground.

What I know is as follows:
x = 100 feet
a = -9.8m/s^2
v_o = 0m/s.

The formula I have tried is 100 = v_o + 0.5(-9.8t). Obviously, it's the wrong formula as I get the wrong answer everytime (the answer I get is 6.1 or 6.2 seconds, can't remember at the moment). I chose that formula because it doesn't require the final velocity at the bottom. I suspect I need to find the final velocity, but I don't know how to go about that.

Any help would be appreciated!

BTW, I registered earlier with the name "Micah," but it never sent me an activation e-mail. Not sure if that's a problem with my e-mail provider (AOL, yay) or the forum. If possible, I'd rather use "Micah" rather than Micah18, so if the admin could resend the activation e-mail (if that is even possible) that would be nice.

2. Dec 15, 2003

Staff: Mentor

Try this formula: $$x=x_0+v_0t+\frac{1}{2}at^2$$

Be sure to use a consistent sign convention.

3. Dec 15, 2003

Micah18

I will try that tomorrow.

Sorry if I am a bit slow, but what do you mean by consistent sign convention?

4. Dec 15, 2003

Hurkyl

Staff Emeritus
Make sure that either + always means upwards or that + always means downwards.

5. Dec 15, 2003

Micah18

Ah yes, will do.

6. Dec 17, 2003

Wooh

Re: Re: Mechanics difficulties - acceleration

In general, I always find it useful that when g = 9.8m/s^2, and there is no air etc and you just drop something, you can just use.
$$\sqrt{\frac{h}{4.9}}$$
The derivation is pretty simple, but it is a nifty reference when you are in a time bind.

7. Dec 18, 2003

No Name Required

It goes without saying that u must change feet into metres (something u didnt do when posting ure first equation in ure first post)