Accelaration and Free-fall problems

[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...

 

[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 traveled 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 traveled by the two rocks is the same.