Question

A rock dropped from a high platform is moving at 24 m/s downward when it strikes the ground. Ignore air resistance. How fast was the rock moving when it had fallen only one-fourth of the distance to the ground?

Answer 1

Which of these equations do you think we should use?

http://en.wikipedia.org/wiki/SUVAT_equations#SUVAT_equations

Hint: We have s, g and u.

Answer 2

s=ut + 1/2 *a * t ^2

Answer 3

Would be my guess

Answer 4

There are a few steps, but not hard:

If you use

v=u-gt

You can find out how long it took to hit the ground

Then you can use

s=(1/2)(u+v)t

to find out how far the rock actually fell.

Take (1/4) of the value of s

Then find out how long it would take to fall this distance using

(s/4)=ut-(1/2)gt^2

Then use

s/2=(1/2)(u+v)t

to find out how fast it was going.

Answer 5

I meant s/4 where I have s/2 above, sorry about that.

Answer 6

This is because we are interested in 1/4 the distance not 1/2 the distance.

Answer 7

Alright, so the first part is 2.45 seconds, I think. 24 = 0 - 9.8t

24 = -9.8t
-2.45 = t (but time obviously can't be negative)!

Look good so far?

Answer 8

But velocity is negative because downward is negative. Also note that s is negative, because s is going down.

Looks good so far.

Answer 9

Note that g was also negative

Answer 10

Ah, okay

Next part...

0.5 * 24 * -2.45 = s = -29.4

-29.4 * 1/4 = -7.35

Answer 11

-7.35 = 0 * -2.45 - 1/2 (0+24) * -2.45

Answer 12

Wait... I messed something up :/

Answer 13

I'm confused again...

Answer 14

You are here: You have s/4 and u=0, g, now solve for t:

(s/4)=ut-(1/2)gt^2

Answer 15

Oh, Okay!

So it should look like this:
-7.35 = 0 * t - 1/2 (0+24) * t

Right?

Answer 16

Wait

Answer 17

That should be 9.8 * t^2 at the end

Answer 18

yes

Answer 19

So the next part is -7.35 = -1/2 * 9.8 * t^2

Answer 20

yes, so
$$
t=\sqrt{\cfrac{2\times 7.35}{9.8}}~seconds
$$

Answer 21

So t = 1.22

Answer 22

I'll let you check calculations, that's the easy part. I'll check formulas.

Answer 23

Sounds good :)

Answer 24

s/4 = (1/2) (u+v) t solving for v

Answer 25

yes

Answer 26

7.35 = 1/2 * v * 1.22

Answer 27

Solve for v and you are done!

Answer 28

So v = 12! I think haha

Answer 29

That was a long one. Thank you so much for your help :)

Answer 30

No problem, I did check that last calculation and got the same. But be sure to double check all your numbers. Formulas are okay.

You're welcome.

Answer 31

Is there any chance you could offer me some advice on two other problems? You won't have to walk me through those, I just want an idea of how I might solve them.

Answer 32

Here is the first one:

A roller-coaster car starts at a height of 90 m at Point A and is moving at a speed of 20 m/s on a later hill at Point C. Assuming friction does not affect the motion of the car, what is its height at Point C?

And here is the second one:

A 0.145 kg baseball is thrown at launch angle of 30° and strikes the ground at 18 m/s. How fast would it be moving when it reaches the ground if its launch angle were 45°? Ignore air resistance.

Answer 33

@wolfe8 Is there any chance you can point me in the right direction with these two questions? You don't need to help much, but I would appreciate any help :)!

Answer 34

For the new questions, can you make new posts for them?