by Jedi Knight Dustin Glazier, Dark Lord of the Sith Josh Patrice,
and the Notorious Gangster Slug Brian the Ward(d), January 99

- Table of Contents
- Formulas
- General Problem Solving Strategy
- Example Problem #1
- Example Problem #2
- Sample Problems
- Solutions to Sample Problems
- Go back to Tutorial Page
- Star Wars

Here's the quantities you can know. Due to the nature of these
problems, you need to look at displacement and velocities in both the horizontal
and vertical planes:

- X Displacement
- t Time
- Vo Initial velocity
- Va Average Velocity
- V Final Velocity
- a Acceleration
- q The angle
- g gravity

- And here are the formulas that we have so far:
- Read the problem.
- Go through the problem and figure out what is given or implied
- Set up a horizontal and vertical table*
- Find any formula that will allow you to calculate
- Add what you just found in the last step to your list of knowns.

1.V = Vo + at

2.Va = (Vo + V)/2

3.X = Vat

4.X = Vot + 1/2at2

5.2aX = V2 - Vo2

Here they are divided into there directional components:

(A subscripted x denotes horizontal and a y denotes vertical to the
respective quantaties)

Horizontal:

V_{x }= V_{xo}

x = V_{x}t

Vertical:

y = ^{1}/_{2} gt^{2 } (on the cliff
problems only)

V_{y }= V_{yo }- gt

y = V_{yo}t - ^{1}/_{2} gt^{2}

V_{y}^{2} = V_{yo}^{2} - 2gy

Go back to: Table of Contents

*What does this mean? What is this? How do I do it, and why? And why is my body growing hair where there was none before? Well, I'll tell you. Because most of the quantities in projectile motion problems (displacement and all the velocities) have both horizontal and vertical components, then it is extremely useful to set up a table to compare the values. Here's the Brian Ward patented method for making one:

Horizontal | Vertical | |

Displacement | x | y |

Initial Velocity | V_{xo} |
V_{yo} |

Average Velocity | V_{a} |
V_{a} |

Final Velocity | = V_{xo} |
V_{y} |

Time | t | t |

Acceleration | a | g |

Things to notice:

Time remains constant for both the horizontal and vertical side. After all, you're just looking at up and down and left and right, not some other wacky dimension.

There is no separate category for the final horizontal velocity because,
in these problems, the horizontal velocity remains constant.

More general problem solving tips for ya'll:

Make a list, and identify the quantities you know.

anything that you don't know, and apply it.

Check to see if you have found the answer. If not, repeat the previous two steps until you are done.

Here's how I generally work a projectile motion problem:

There's actually two different types of problems which fit under the category of projectile motion. The first, and easier type, is called a cliff problem. In it, an object rolls, is pushed, or somehow goes off of a cliff. It has an initial horizontal velocity, which sometimes is given to you (and sometimes you must find it- hahaha). Since it went straight off a cliff, it generally has no initial vertical velocity. Instead it accelerates downward at the rate of 9.8 meters per second squared. As a quick refresher on acceleration, that means that for every second which passes, the velocity downward increases by 9.8 m/s. Cool, huh? Therefore knowing time and the acceleration can tell you the velocity at any point.

And that brings me to what I consider to be the key to solving these
problems: finding out time. What I mean by time is the time
it takes for the object to hit the ground after going off the cliff.
Time is the one element shared by both the horizontal and vertical sides,
so by finding it out you can often bridge the gap between the two and discover
the unknowns. There are a few ways to determine this. If you
start off the problem being given the distance from the top of the cliff
to the ground, then you can easily figure out the time it takes to hit
the ground. Since the initial velocity is always 0 and you know acceleration
(referred to as g for gravity in the formulas) is 9.8, then you just use
the equation

y = V_{yo}t - ^{1}/_{2} gt^{2}
and away you go. Of course, in some instances you are not given the
distance to the ground but instead you know how far horizontally the ball
travels before reaching the ground. That's even easier to use to
determine the time, for if you know the horizontal velocity you just divide
that by the distance traveled, and you get the time! Yay!!

Sometimes, you need to find the overall speed of the object, which involves
both the horizontal velocity and the vertical velocity. The way to
do this is to find out the velocities of both of those separately, and
then use vectors (remember those fun things). Refer to the earlier
work done on vectors to allow you to combine the velocities and come up
with a total one (remember the chant: SPEED IS THE HYPOTENUSE!!).

While there are other things you have to do with cliff problems, they are fairly rudimentary and shouldn't require anything more than a perfunctory understanding of the equations given. A tad bit more difficult, however, are the ARC problems. The only difference is that they start from the ground, go up, stop at a point, and then come down. It is important to realize that after the object reaches the top of the arc, the rest of the problem is exactly like a cliff problem. So there really isn't much difference is how to do the work, there's just more of it. Remember, that at the top the vertical velocity is 0. And also make sure to differentiate negative and positive velocity while using the equations.

Good luck, my students. With hard work and perseverance, you will prevail.

Go back to: Table of Contents

A cannon ball (soaked in napalm) is shot off a 100-m-high cliff in order to provide an example for our first problem. The ball has an initial horizontal velocity (V

Here is what you start with:

Horizontal | Vertical | |

Displacement | 90.0 m | 100 m |

Initial Velocity | ? | 0 m/s |

Final Velocity | ? | ? |

Acceleration | 0 | 9.8 m/s/s |

Time | ? | ? |

So, if you look at the formulas, find the one with what you have. Since we're trying to find time use the one that has y and g. Have you found it? We'll wait a while longer. Done yet? No? Okay, we're moving on.

so

y =^{ 1}/_{2} gt^{2}

This looks rather messy, lets clean it up to solve for time:

t = (2y/g)^{(1/2)}

when numbers are inserted:

(200 m/9.80 m/s^{2})^{(1/2) }= 4.52 s

you've learned time congratulations

D: didn't we learn time
and just tell them what it was?

B: yeah really!

JP: by that theory, didn't
Brian learn physics and just tell us what it was?

D: fair enough.

now we have this:

Horizontal | Vertical | |

Displacement | 90.0 m | 100 m |

Initial Velocity | ? | 0 |

Final Velocity | ? | ? |

Acceleration | 0 | 9.8 m/s/s |

Time | 4.52 s | 4.52 s |

So again you look at the formulas. What will you use? Here's a hint:

x = V_{xo}t

after some cleaning:

V_{xo }= x/t

Here's the numbers:

V_{xo }= 90.0m/4.52s = 19.9 m/s.

Now we know these quantities:

Horizontal | Vertical | |

Displacement | 90 m | 100 m |

Initial Velocity | 19.9 m/s | 0 |

Final Velocity | 19.9 m/s | |

Acceleration | 0 | 9.8m/s/s |

Time | 4.52 s | 4.52 s |

Now we know everything

Since the question asked for initial velocity, you would need to round
to the

number of significant digits you have. (3 in this case)

so initial velocity = 19.9m/ s.

Go back to: Table of Contents

An Australian football is kicked at an angle q

Here's what we know:

Horizontal | Vertical | |

Displacement | ? | ? |

Time | ? | ? |

Initial Velocity | V_{o} cos 37.0^{o} |
V_{o} sin 37.0^{o} |

Average Velocity | ? | ? |

Final Velocity | ? | ? |

Acceleration | 0 | 9.8 m/s/s |

Look at the formulas again, they are taunting you aren't they? This one tends to do what we need it to, which is set-up the problems at hand by using these we can find initial velocity for both the horizontal and vertical:

V_{xo }= V_{o} cos 37.0^{o }= (20.0 m/s)(0.799)
= 16.0 m/s

V_{yo }= V_{o} sin 37.0^{o }= (20.0 m/s)(0.602)
= 12.0 m/s.

(a) The max height is attained where V_{y} = 0, this occurs
when t = V_{yo}/g, ergo (12.0 m/s)/(9.80 m/s^{2}) = 1.22
s.

Time for a new formula:

y = V_{yo}t - ^{1}/_{2} gt^{2}
^{ }= (12.0 m/s)(1.22 s) - ^{1}/_{2}(9.80
m/s^{2})(1.22 s)^{2}
^{ }= 7.35 m.

(b) To find the time it takes for the ball to return to the ground, we use the following equation and set y = 0 (for ground level).

y = V_{yo}t - ^{1}/_{2} gt^{2}

0 = (12.0 m/s)t - ^{1}/_{2}(9.80 m/s^{2})t^{2}

thus,

t = ^{2(12. m/s)}/_{(9.80 m/s}^{2 }_{)}
= 2.45 s

you might have found t = 0, this is a solution, congratulations, but it is wrong, it is 0 when the initial point y is zero.

(c) The total distance traveled horizontally is found by applying the
following equation, remembering that a = 0, V_{xo} = 16.0 m/s:

x = V_{xo}t = (16.0 m/s)(2.45 s) = 39.2 m

(d) At the max height there is no vertical component to the velocity,
only horizontal. so v = V_{xo} cos 37.0^{o} = 16.0 m/s.

(e) The acceleration vector is always 9.80 m/s^{2} downward.

Go back to: Table of Contents

The answers to each problem follow it in parentheses. They also link to a solution to

the problem. Try the problem, check your answer, and go to the solution if you do not

understand.

a) Draw the initial velocity vector. Find the initial speed. b) For what time is the Quantum in the air? c) How far does it go in this time d) should the Quantum be painted with the characatures of Scott Bakula and Dean Stockwell?

(67.6 m/s, 10.8 s, 454 m, hel* yess!!!)

( 3.96 m, 12.2 m/s, 7.1 m)

Brian leaves the edge of a 12m tall cliff with a horizontal velocity
of 4.8 m/s. What time is he in the air? How far from the base
of the cliff does he land? What is his velocity upon impact with
the ground in terms of x and y components?

(1.56 s, 7.5 m, 4.8 m/s x + -15.3 m/s y)

(590 m/s, 413 m/s, 84.3 s, 49,700 m)

Go back to: Table of Contents

5. Brian's girlfriend jumps off the edge of a cliff and hits the water 1.5 seconds later, about 4.5 m from the edge of the cliff. What height was the cliff? With what speed did she leave the edge? Why did she jump?

(11 m, 3 m/s, she was dating Brian)

(2.07 s,1.64 m/s, because he's focused)

(14.9 m/s, 23 m/s, 4.7 sec, 70 m,76.6 m, no he's tied with Mike D and MCA)

(a) 27.1 m/s x + 20.5 m/s y,b) 4.1 s, c) 113 m, d) 27 m/s, e) 27.1 m/s x + 14.9 m/s y, f) 31 m/s, g) for the sake of our grade, Yeas!)

Go back to: Table of Contents

a) Draw the initial velocity vector. Find the initial speed. b) For what time is the Quantum in the air? c) How far does it go in this time d) should the Quantum be painted with the characatures of Scott Bakula and Dean Stockwell?

(67.6 m/s, 10.8 s, 454 m, hel* yess!!!)

Here is what you start with:

Horizontal | Vertical | |

Displacement | ? | ? |

Time | ? | ? |

Initial Velocity | 42 m/s | 53 m/s |

Final Velocity | ? | ? |

Acceleration | 0 | 9.8 m/s/s |

Ok first close your eyes. Now picture the initial velocity vector.
You
got that, good now that's your picture. Now for physics! For the
speed, we again visit Pathagorean. 53^{2} + 42^{2} = initial
speed^{2}. So initial speed = 67.6 m/s. For time,
V_{y }= V_{yo }- gt. Plug it in, plug it in.
You know the final vertical velocity is the opposite of the initial vertical
velocity. so that means the -53 = 53 - (9.8 m/s/s)t. So math yields t =
10.8 s, great. We can now find the distance it went using V_{x}t
= x. So (42 m/s)(10.8) = 454 m

The Quantum would look really f*****' cool if it had a b*$*&%@ pictures of both Dean the j#$& Stockwell and that other guy, a.k.a. Scott Bakula.

Go to: Problem Formulas
Table of Contents

Dustin leaps from the edge of a cliff with a velocity of 3.3 m/s
horizontally. He hits the water 1.2 seconds later. How far
out does he land? What is his speed when he hits? How high is the
cliff?

( 3.96 m, 12.2 m/s, 7.1 m)

Here is what you start with:

Horizontal | Vertical | |

Displacement | ? | ? |

Time | 1.2 s | 1.2 s |

Initial Velocity | 3.3 m/s | 0 |

Final Velocity | 3.3 m/s | ? |

Acceleration | 0 | 9.8 m/s/s |

Overall speed | ? | ? |

OK let's go. Use V_{x}t = x plug in de stuff. Get
(3.3 m/s)(1.2 s) = 3.96 m. Good. Now we must find the velocity of
y first. Vy = at so (9.8)(1.2 s) = 11.76 m/s vertically. Use
the force, no use the pathagorean theorummm, so 3.3^{2} + 11.76^{2}
= the hypotneuse^{2} which we know as SPEED! So the hypotenuse
equals (square root)150 or 12.2! Finding the height is easy.
Use y = 1/2gt^{2} and voila u get 7.1 m.

Go to: Problem Formulas Table of Contents

Brian leaves the edge of a 12m tall cliff with a horizontal velocity
of 4.8 m/s. What time is he in the air? How far from the base
of the cliff does he land? What is his velocity upon impact with
the ground in terms of x and y components?

(1.56 s, 7.5 m, 4.8 m/s x + -15.3 m/s y)

Here's what you start with

Horizontal | Vertical | |

Displacement | ? | 12 m |

Time | ? | ? |

Initial Velocity | 4.8 m/s | 0 |

Final Velocity | 4.8 m/s | ? |

Average Velocity | ? | ? |

Acceleration | 0 | 9.8 m/s/s |

Well, this is easy, errr something. let's find the time using:
y = 1/2 gt^{2} ok. But we'll use it looking like this: t = (square
root) 2y/g

t = (square root) 2*(12m)/(9.8 m/s/s)

yielding: t = 1.56 s

Now onto how far he goes, we use the formula V_{x}t = x so
(4.8 m/s)(1.56) = 7.5 m SWEET!

The Vector components seem hard, hehehe hard. But they're not,
use the brain.

and then put it in to the form: (x velocity here) m/s x
+ (y velocity here) m/s y

we know the x velocity = 4.8 and the y velocity = a(cceleration)t(ime)
or (9.8 m/s/s)(1.56s) which equals 15.288

since it goes down it is negative thus it is 4.8 m/s x + -15.3 m/s
y.

Go to: Problem Formulas
Table of Contents

(590 m/s, 413 m/s, 84.3 s, 49,700 m, Red)

Horizontal | Vertical | |

Displacement | ? | ? |

Time | ? | ? |

Initial Velocity | ? | ? |

Final Velocity | ? | ? |

Accleration | 0 | 9.8 m/s/s |

Colour | yeah | uh huh |

You know the speed, which is both horizontal and vertical. Use
vectors to find horizontal and vertical initial velocities.

cos 35 times 720 = horizontal velocity = 589.79 m/s

sin 35 times 720 = vertical velocity = 412.98 m/s

time -

412.98 = -412.98 - 9.8t

t = 825.95/9.8

t = 84.28 s

horizontal displacement is horizontal velocity times time

(589.79) 84.28 = 49707.5 m

As far as your favorite color, I guess that is a matter of personal
choice. I think that red is a fine personal choice, so I chose it. Don't
you think you ought to choose red too?

Go to: Problem Formulas
Table of Contents

Brian's girlfriend jumps off the edge of a cliff and hits the water
1.5 seconds later, about 4.5 m from the edge of the cliff. What height
was the cliff? With what speed did she leave the edge? Why did she
jump?

(11 m, 3 m/s, she was dating Brian)

Here we go

Horizontal | Vertical | |

Displacement | 4.5 | ? |

Time | 1.5 s | 1.5 s |

Initial Velcocity | ? | 0 |

Final Velocity | ? | ? |

Acceleration | 0 | 9.8 m/ |

Girlfriend | never | whenever I can |

Use the formula y = V_{yo}t - ^{1}/_{2}
gt^{2} since initial vertical velocity is 0, then you can quite
easily plug in the time and g and find out y, the distance from the top
of the cliff to the ground. By doing that (remember to take the square
root of the answer on the left side because t is squared, my darlings)
you should get an answer of 11 meters (with those darned sig figs).
To find the initial horizontal velocity is a sinch from here on out.
Divide the distance traveled horizontally (4.5 m) by the time (1.5 s) and
you get 3 m/s.

And, for the easiest portion of the question, she flung herself into
the oblivion of oblivion because she was going out with me. That's
right. Me, Notorious Gangster Slug Brian the Ward(d). You know,
this joke really hurts me. It really does.

Go to: Problem Formulas Table of Contents

(2.07 s,1.64 m/s, because he's focused)

Horizontal | Vertical | |

Displacement | 3.4 m | 21 m |

Time | ? | ? |

Initial Velocity | ? | 0 |

Final Velocity | ? | ? |

Acceleration | 0 | 9.8 m/s/s |

Time first mon cherie, now do the same equation from number 1 above.
Why is there 1, 2, 3, 4, and then 1, 2, 3, 4 again, I don't know? In any
case use the second 1 not the first second, and definitely not the first
first. y = V_{yo}t - ^{1}/_{2} gt^{2}
remember the last problem, or did you even bother doing it.

JP&D: we wouldn't bother doing it.

NGSBW (Notorious Gangster Slug Brian the Ward(d)): Oh I compreh

So plug in the y and g and let's see where we go. 21 = 1/2 (9.8)
tThis gives you a t of 2.07 seconds. For horizontal velocity, you
use the equation X = Vt. 3.4 = 2.07v

V = 1.64 m/s

Now to by far the most important part of the question. Why is Leo so much of a frenchie? Think tanks have debated long and hard on the subject. There are many answers to this question. The most obvious is that he is from france. But I fear that this may be too simple of an explanation. One must also consider that he already has over 1000 words on his extended essay, however, he will need to give up shortly to prove this theory correct. Another possibility is that he could have fallen on his head as a young child. All we know is that everyone loves Princess Leo deep down (B: well, in a figurative way. JP: Which means not at all).

Go to: Problem Formulas
Table of Contents

(14.9 m/s,23 m/s,4.7 sec, 70 m,76.6 m, no he's tied with Mike D and
MCA)

Horizontal | Vertical | |

Displacement | ? | ? |

Time | ? | ? |

Inital velocity | ? | ? |

Final Velocity | ? | ? |

Accleration | 0 | 9.8 m/s/s |

a) B: What do I look like, Picasso? Draw it yourself.
Show it to me later and I'll tell you if its right or not

b) JP: thanks for that thing ghuy. D: why did you say ghuy?
JP: its a regional dialect. anyway, you gots ta use da' vectas, remember
the vectas? (B: he means vectors. Reference chapter whatever)

Since um yeah...

In any case, with the angle of projection and the initial velocity (which includes both horizontal and vertical components) you can figure out the separate horizontal and vertical velocities. Let's start with horizontal. Take that handy picture you drew, use it and your knowledge of vectors and this should be simple. The horizontal velocity will be equal to the bottom line of the triangle. To find that, you need to use cosine. The cosine of 57 degrees will be equal to the horizontal velocity divided by the hypotenuse (which we know as 27.4). Make sure you're in degrees, not radians (because I just tried the problem in radians, and let me tell you it does not work!). Multiply the cos of 57 by 27.4 and you get 14.9 m/s, which luckily enough is the answer.

c) Same process for this as in b. But this time, since you're finding the side opposite to the angle, you use sin, not cos. The sin of 57 times 27.4 is 22.9 m/s, which we shorten to 23 for our own reasons.

d) Since we know the vertical velocity, we can now discover the
total time. Use the equation V_{y }= V_{yo }- gt
Since you know that the final vertical velocity will just be the opposite
of the initial velocity (don't ask me why, its just true) you can easily
find time. Initial velocity is 23, while the final will be -23.
So you get

-46 = -gt

And since g is 9.8, that transforms into

46/9.8 = t

t is 4.7 s

e) Now just multiply the horizontal velocity by the time its in the air. 4.7 times 14.9 = 70 m.

f) This problem is quite a challenge. First off, you have
to manipulate some of the formulas you already have. One of the few
things that you know is that at the point of maximum range, y = 0 (since
you're finding out where it'll hit the ground). Plug this into the
formula y = V_{yo}t - ^{1}/_{2} gt^{2 }
and you get 0 = V_{yo}t - ^{1}/_{2} gt^{2}

Solve this expression for t, which will give you the solutions t=0 and
t= (2v_{yo})/g. This makes sense. The ball is level
with the ground at the time it is kicked, and the second equation tells
you the time it'll take for it to reach the ground after the kick.
Since this equals time, you can substitute it into the equation x =
V_{x}t. Multiplying the horizontal velocity by the expression
we got earlier will eventually give you the equation x = (v_{0}^{2}
sin 2q)/g. To get all that
you have to use fancy trigonometric formulas and math (which I hope you
learned- if not complain to your math teachers). If you're confused,
then just trust me. I'm right. Now you have to plug in the
angle which gives one the best range, which would be 45 degrees.
Plug that, the velocity, and g into the equation, and you receive an answer
of 76.6 m. Hope that was fun for ya'll.

Go to: Problem Formulas Table of Contents

(a) 27.1 m/s x + 20.5 m/s y, b) 4.1 s, c) 113 m, d) 27 m/s, e) 27.1
m/s x + 14.9 m/s y, f ) 31 m/s, g) for the sake of our grade, Yessss!)

Horizontal | Vertical | |

Displacement | ? | ? |

Time | ? | ? |

Initial Velocity | ? | ? |

Final Velocity | ? | ? |

Accleration | 0 | 9.8 m/s/s |

You're just finding the vertical and horizontal velocities again, but at the same time. Again, take the cosine of 37 and multiply it by 34 for the horizontal, and the sin of 37 and multiply it by 34 for the vertical.

b). use the same equation as in d) of the last problem.
-41 m/s = 9.8t. t = 4.1s

c). 4.1 times 27.1 = 113 meters.

d). at the highest point, the vertical velocity has to be 0 (because
it has ceased its climb and is about to start coming down). So all
you have is horizontal velocity, which doesn't change one stinking bit.
That's 27.1 m/s

e). This is sorta tricky. You know the y distance, but
need to know the vertical velocity at that moment. You need to use
2 formulas. First off use y = V_{yo}t - ^{1}/_{2}
gt^{2} You know everything except time Find that. Now
you know how long it takes for the ball to reach 10m. Plug that into
the equation V_{y }= V_{yo }- gt With that
you can find the velocity at that exact instance easily. Now you
know the vertical and horizontal velocities (horizontal always stays the
same).

f) Use Pythoragram's theorem (a squared time b squared = c squared). to combine the two velocities into one comprehensive velocity.

Well, now you're done. I hope this experience has served you well. Your life has been enriched by our teachings of:

PROJECTILE MOTION!!!!

Thank you for you time, and good luck. May the force go with you.
.............................................................Always

SHUT UP Dustin!