# Free fall acceleration concept question

1. May 16, 2013

### xzibition8612

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

A student throws a set of keys vertically upward to her sorority sister in a window 4m above. The keys are caught 1.5s later by the sister's outstretched hands. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?

2. Relevant equations

3. The attempt at a solution
(a) 10m/s
(b) -4.68 m/s

I know how to get the solutions by using the constant acceleration formulas, but my question is with (b). If the velocity is negative, that means the keys have to be moving DOWN, because maximum height achieved by the key would be when the velocity is 0. So I'm very confused on this point. How could the key still be going up if its already negative velocity? Someone please point out where my thought process went wrong.

2. May 16, 2013

### ap123

Would would happen if she didn't catch the keys on the way up?

3. May 16, 2013

### xzibition8612

if she didn't catch the keys it would fall down. but i don't get it because that's supposed to happen when velocity is zero, and this -4.68m/s velocity means theoretically the sister shouldn't even be able to touch the key since it would be falling down earlier.

4. May 16, 2013

### ap123

To put it in another way, how long does it take the keys to reach their highest point, ( you can easily get this from your answer to part (a) )

5. May 16, 2013

### xzibition8612

v= v0+at

=> 0 = 10 + (-9.8)t

t = 1.02 s

It takes 1.02s for the key to reach its apex. Wow this contradicts the 1.5seconds given by the problem. What's going on!!?

6. May 16, 2013

### CAF123

If the keys are thrown with an intial velocity approx 10m/s, at what height are they after 1s? And if the woman catches them at a height 4m at t=1.5, then where is the keys headed?

Last edited: May 16, 2013
7. May 16, 2013

### ap123

So, what happens to the keys after this?