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...

 

[tiny-tim]

Welcome to PF!

Hi Raheelp! Welcome to PF!

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)

If up is positive, then it should be +8, not -8

Question B:

0 = v02 + 2 * -6.7 * 3.8

You need two equations, one for the reaction time (zero acceleration), and one for the braking.

Question C:

Show us what equations you have got, and then we'll know how to help!

 

[Raheelp]

Hi Raheelp! Welcome to PF!


If up is positive, then it should be +8, not -8



You need two equations, one for the reaction time (zero acceleration), and one for the braking.


Show us what equations you have got, and then we'll know how to help!

OK ty on question A it was obvious..

Question B I am trying right now.

The equations we use are the usual three of motion and utilize average speed, average velocity.

 

[Raheelp]

So for B:

3.8 m = segment 1 d + segment 2 d ?

 

[tiny-tim]

So for B:

3.8 m = segment 1 d + segment 2 d ?

Yup!

Show us what you get. ​

 

[Raheelp]

Yeah I can't do B, I fail at this.

So far I have this:

For segment 1 I have

a = 0
t = .45
d = x
v0 = ?
v = ?

Segment 2

a = -6.7
t = ?
d = 3.8 - x
v0 = ?
v = 0

Do I solve for v0 in segment two, and put that in segment 1 to find d... ?

 

[tiny-tim]

Do I solve for v0 in segment two, and put that in segment 1 to find d... ?

Yes.

(but you can't call both of the distances 3.8 - x, can you? make the first distance x )

(oh, and write v_i if you're writing v_f … write v₀ only if you're writing v₁ )

 

[Raheelp]

Now I'm confused with what formula I can use.

I think I'll have to get walked through the problem at the workshops on Monday.

Thanks guys... Sorry I couldn't do anything with your expertise.