View Full Version : vertical motion upward

MatthewMM

05-11-2006, 11:26 PM

I am working on a problem which I'm confused about. I think I know how to do vertical motion downward but I'm not sure what to do about upward. The problem is : A person standing on a creek bed throws a rock straight up at a speed of 64 feet per second.

1) When will the rock reach a maximum height?

2) When would the rock hit the water again?

I know the formula for vertical motion is h=-16t^2+vt+s

When I did a problem throwing something down from a bridge I put the height at o because I was throwing downward. Could someone please help me on how I get started with the equation going upward?

tkhunny

05-12-2006, 12:16 AM

You need to put zero (0) where ever it needs to be. Generally, that would be given in the problem statement. In this case, you're in a creek bed. That's a good place for zero (0). Take v = +64 ft/sec and s = 0 ft

MatthewMM

05-12-2006, 01:27 AM

Would I still use the same formula of h=-16t^2+vt+s and put 0 fors and 64 for v. When I work it out I get t=0 or t=4. But the answer to the question is when will it reach a maximum height. Which answer would that be. Also the next question is when would it hit the water again. How do I use that same formulas for that?

tkhunny

05-12-2006, 07:12 AM

If your only force is gravity, you have the right formula.

Good, you have h(0 sec) = 0 ft and h(4 sec) = 0 ft. You should hear a splash at t = 4 sec, right?

An important property of these things is the maximum is ALWAYS in the same place, exactly between the two zeros. See if you can get a value greater than h(2 sec).

MatthewMM

05-12-2006, 04:59 PM

okay so I've got that it will take 4 seconds to reach a maximum height but what do I do to figure when the rock will hit the water again?

stapel

05-12-2006, 05:28 PM

Hint: What is the height (above the water) when the rock hits the water?

Eliz.

MatthewMM

05-12-2006, 07:38 PM

The height when the rock hits the water would be 0 but what do I use for s and do I multiply the 64 feet per second by 4 to get the v?

MatthewMM

05-12-2006, 08:07 PM

Would it take 4 seconds to hit the water just like it took 4 seconds to reach maximum height? It seems that it would come down quicker!

tkhunny

05-12-2006, 10:44 PM

Let's review...

Good, you have h(0 sec) = 0 ft and h(4 sec) = 0 ft. You should hear a splash at t = 4 sec, right?

Please don't use t = 4 sec for anything else.

MatthewMM

05-12-2006, 11:43 PM

I'm sorry I'm so confused on this. The first problem I used the formula of

h=-16t^2 + vt = s and I put h= -16t^2 + 64t +0. After working it out, I came up with t =0 and t =4 Therefore I came to the conclusion that it would take 4 seconds to reach maximum height at a speed of 64 feet per second. Is that not right? Then I didn't know how to get when the rock would hit the water again. I've been pondering this problem for two days! At this point I'm very confused! Thanks for your help!

Mrspi

05-13-2006, 10:44 AM

okay so I've got that it will take 4 seconds to reach a maximum height but what do I do to figure when the rock will hit the water again?

I have two thoughts here:

1) You didn't read the previous responses carefully.

2) Perhaps you are "overthinking" this problem.....making it harder than it has to be.

The height is 0 (the rock is at the surface of the creek) TWO times....when t = 0, and when t = 4. That is, at 0 seconds, and again at 4 seconds. The second time that the height is 0 would be when the rock hits the water again!! So, when does the rock hit the water again?

One of the previous responders pointed out that the maximum height (the high point you're looking for) occurs when t is precisely halfway between the two values of t for which h = 0.....that is, between t = 0 and t = 4. What's halfway between 0 and 4? Use this value for t in your equation, and evaluate h. That will be the maximum height reached by the rock.

I hope this helps you.

tkhunny

05-13-2006, 11:00 AM

This is a Uniform Acceleration problem. Your thoughts about 64 ft/sec are a bit off. Gravity slows down the object and speeds it up later. It is going 64 ft/sec only for a moment, then slows, and slows, and slows, until the maximum height, when the vertical velocity is zero (0) ft/sec. Gravity then begins to accelerate the object back down at an increasing rate. Only at the moment of splash down does it again achieve 64 ft/sec. Speed (velocity) is constantly changing. That is what makes this problem interesting.

MatthewMM

05-13-2006, 02:13 PM

So if I half the 0,4 then it will reach maximum height at 2 seconds and splash back down at 4 seconds?! Since I don't have to put what the height is then do I still need to put 2 in the t place. This has gone way over my head! Thank youo for being so patient with me!

tkhunny

05-13-2006, 04:45 PM

This has gone way over my head!Where else would you expect a rock to go when you throw it in the air? :lol:

You throw it upward.

It comes back.

In the middle, it reached maximum height.

Of course, this ignores friction, wind, intervening geographical features, birds, and various other things.

t (time) is the independent variable.

Pluf it in (y = f(t)) to calculate the value of the dependent variable (height).

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