# Horizontal and Vertical components of falling objects

1. Dec 14, 2008

### K0tsk

Seen as i had difficulty understanding this little topic and with my AS level exam soon, i figured i should find somewhere to ask without having to wait while class, so here goes...

I had some questions to do (but this isn't about homework!!) i couldn't seem to be able to get the right answer and so i figure maybe my approach is wrong.

I understand that simple the horizontal component will be the speed x time, but everytime i try find the vertical component, at say X time (in a SHOW THAT question) i can't seem to be able to find the right answer, can anybody explain?

btw i did search for this topic, but no-one seemed to cover the area of it that im stuck with

thanks

2. Dec 14, 2008

### CompuChip

Welcome to PF, K0tsk.
The important point for such questions is to always remember that you can treat the horizontal and vertical directions independently. Since horizontally, there is no acceleration, you can use the corresponding formulas for such a motion (ax = 0, vx = constant, x = v t). The vertical motion is of course accelerated due to the planet pulling on the object, and we usually call such an acceleration g (on earth, g is something like 9,8 - depending on where you live; on the moon it's only 1,6..something). So for the vertical part, you should use the formulas corresponding to such motion (a = -g, $\Delta v_y = - g t$, $\Delta y = - \frac12 g t^2 + (v_y)_0 t$).

Usually a falling object-problem consists of
a) thinking what the initial variables are (what is the initial velocity in the x direction and the y direction, the initial height)
b) solving one of the formulas for the falling time
c) plugging the time into the other formulas to find a distance

For example, if they ask you how far from the cliff your thrown rock will land, you first solve the vertical formula for the falling time, then plug that into the horizontal formula. If they ask if Superman will make it across the gap, you first solve the horizontal formula for the time he must be in the air and check with the vertical formula if he will still have positive height then (or vice versa: check how long he can stay in the air and how far he will travel at most).

So that's the general story, if you want something more specific maybe it's useful to work through a problem.

3. Dec 14, 2008

### K0tsk

Ah, i see now. The formulas are pretty confusing though. I had a question which was first to obtain an expression for v in terms of u, a and s from;
v=u+at
s=(u+v)t/2
which i wasn't sure if this is what i needed to do the next part, which was..
Ball kicked from cliff with horizontal velocity 5.6...
Show that after 0.90s vertical component of velocity is 8.8 ms-1

This got me confused, tried a variation of formulas from other notes and still couldn't get the 8.8 !

4. Dec 14, 2008

### CompuChip

That's not really going to work. You will have to learn to think toward the answer. It's generally a good idea to teach yourself some scheme like the one below. It may seem like a waste of time to first go through all those steps, but after some practice you
will find that thinking about the problem first will save you time in the end. Usually, if people cannot solve a question, it is because they get themselves caught up in mathematics (like calculating everything they can find) without any purpose, just hoping the answer will drop out somewhere by luck.

So an example of such a solving scheme (applied to your question) would be:

1) What is asked for? (Write down a relevant formula, usually that will already get you marks)
- A vertical component of the velocity. Vertically, there is an acceleration, so the correct formula is v = u + a t.
- What do all the variables mean and what are their units?
- In this formula, what variable do we want to solve for?
2) What information do I need?
- You want to know the velocity, v. So you will need to know v(0), a and t. What does the question explicitly give? What is implicitly given? Is there anything you don't know, that you will have to calculate from another formula? (If so, apply the scheme to those quantities as well).
3) After setting out your strategy as in 2), do the calculation (plug in the numbers, apply your math skills to solve for the unknowns).
4) Evaluate: is your answer reasonable? If the stone falls for less than one second, for example, can its vertical velocity be negative? Can it be more than 10 m/s?

I left some questions in italic font for you

5. Dec 14, 2008

### K0tsk

I worked through that question, using that formula v=u+at, problem is though i thought u is the initial velocity which would be 5.6 wouldn't it? I did get 8.8 from 9.81 x 0.90 though.

Also, do you have any advice for being able to handle the kinetmatic equations ? I get them in the front of the exam but is it best to learn what each letter means and then use appropriately in the exam?

6. Dec 14, 2008

### Staff: Mentor

No, the relevant initial velocity is the vertical component, which is 0. Recall the first thing that CompuChip said: Treat vertical and horizontal motion independently.
Well, yeah. (If you don't know what the letters mean, of what use are the formulas?)

You might profit from these links:
For kinematics in general: http://hyperphysics.phy-astr.gsu.edu/Hbase/mot.html
For projectile motion: http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html

7. Dec 14, 2008

### K0tsk

I see what you mean. Ill go back and take a look at some questions.

Cheers

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