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.

 

[tomcue]

As a scientist, it is important to always double-check your calculations and make sure your units are consistent. For Question A, I would suggest using the correct units for acceleration (m/s^2) and time (s) in your equation. It should be -8m = 19t - 1/2 * 9.8 * t^2. Also, make sure to use the correct formula for free-fall, which is d = v0t - 1/2 * gt^2.

For Question B, your approach seems correct. To find the fraction of the 3.8m due to reaction time, you can use the equation d = v0t + 1/2 * at^2, where d = 3.8m, a = -6.7m/s^2, and t = 0.45s. Solve for v0 and then divide it by 3.8m to find the fraction.

For Question C, you can use the equations of motion to solve for the height of the cliff. The first stone has a constant downward acceleration of -9.8m/s^2 and travels for 2.2 seconds. The second stone has an initial velocity of 49m/s and travels for an unknown time until it reaches the water at the same time as the first stone. Set up two equations, one for the first stone and one for the second stone, and solve for the height of the cliff.