# Accelaration and Free-fall problems

1. Sep 17, 2009

### Raheelp

Question A:

This one is easy but my answer is coming out wrong, maybe my signs are wrong. The question is as follows:

A ball is launched directly upward from ground level with an initial speed of 19 m/s.

How many seconds after launch is the ball 8 m above the release point?

I set it up like this but it keeps coming wrong:

-8 m = 19t - 1/2 * 9.8 * t2(squared)

Question B:

This one is driving me nuts, I swear I have it but I dunno.

A typical automobile under hard braking loses speed at a rate of about 6.7 m/s2; the typical reaction time to engage the brakes is 0.45 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 3.8 m.

(a) What maximum speed does this imply for an automobile in this zone?

I used V2 = V02 + 2ad and it looks like:

0 = v02 + 2 * -6.7 * 3.8

The answer is in m/s, convert that to mi/h to get 15.98 mi/h. The other part to the question is:

(b) What fraction of the 3.8 m is due to the reaction time?

Which I believe I need part 1 for.

Question C:

This question I have no clue where to start.

At t = 0, a stone is dropped from a cliff above a lake; 2.2 seconds later another stone is thrown downward from the same point with an initial speed of 49 m/s. Both stones hit the water at the same instant. Find the height of the cliff.

Any help at all will be appreciated greatly...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 17, 2009

### Anden

The first question you have the right equation but why is d = -8? Remove the minus sign, then everything is fine.

For the second: The distance d travelled is equal to the reaction distance + the breaking distance. You'll need d = Vo t + 0,5at^2 and V = Vo + at for this.

For the third: Use the distance formula, d = Vo t + 0,5at^2. I only solved this graphically because I was lazy, but know that the distance travelled by the two rocks is the same.