Projectile Motion, or how to HUCK

Table of Contents

• 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:

 

 

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_xt

Vertical:
y  = ¹/₂ gt²  (on the cliff problems only)
V_y = V_yo - gt
y  = V_yot - ¹/₂ gt²
V_y² = V_yo²  - 2gy

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General Problem Solving Strategy:

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1. Read the problem.
2. Go through the problem and figure out what is given or implied
3. Set up a horizontal and vertical table*

 *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	Vxo	Vyo
Average Velocity	Va	Va
Final Velocity	= Vxo	Vy
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.

4. Find any formula that will allow you to calculate

anything that you don't know, and apply it.
5. Add what you just found in the last step to your list of knowns.

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_yot - ¹/₂ gt² 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.


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Example problem 1

The cliff of terror

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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_xo), but no vertical velocity. It strikes the ground 90.0m from the base of the cliff. What was the initial velocity (speed when thrown)?

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 gt²

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/s²)^(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_xot

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.

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Example problem 2

The arc de tedious

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An Australian football is kicked at an angle q_o = 37.0^o with a velocity of 20.0 m/s. Please find (a) the max height, (b) the time in air before striking the ground, (c) the horizontal distance traveled, (d) the velocity vector at the max height, and (e) the acceleration vector at max height.

Here's what we know:
 

	Horizontal	Vertical
Displacement	?	?
Time	?	?
Initial Velocity	Vo cos 37.0o	Vo sin 37.0o
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/s²) = 1.22 s.
Time for a new formula:

y = V_yot - ¹/₂ gt²
    = (12.0 m/s)(1.22 s) - ¹/₂(9.80 m/s²)(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_yot - ¹/₂ gt²
0 = (12.0 m/s)t - ¹/₂(9.80 m/s²)t²

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_xot = (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/s² downward.
 

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Sample Problems

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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.
 
 

1.

Brian's car, the Quantum, leaves the ground with an initial vertical velocity of 53 m/s and a horizontal velocity of 42 m/s.
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!!!)

2.

Dustin leaps from the edge of a cliff with a velocity of 3.3 m/s horizontally.  She hits the water 1.2 seconds later.  How far out does she land? What is her speed when she hits?  How high is the cliff?
( 3.96 m, 12.2 m/s, 7.1 m)

3.

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)

4.

A Dustin is fired at an angle 35^o above the horizontal at a speed of 720 m/s.  a) Draw a picture of the initial velocity vector.  b) What is the horizontal velocity?  c) What is the initial vertical velocity component? d) What time will the shell be in the air?  e) What distance will it go in that time?
(590 m/s, 413 m/s, 84.3 s, 49,700 m)

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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)
 
 

6.

Leo wants to jump into a big vat of frenchie cheese from a cliff that is 21 m tall.  There are rocks that stick out 3.4 m from the base of the cliff.  What time will he be in the air?  What must his speed be in order to clear the rocks? Why is he soooo frenchie?
(2.07 s,1.64 m/s, because he's focused)

7.

Adrock leaves the ground at a speed of 27.4 m/s at an angle above the horizontal of 57o.  a) Draw a picture of the initial velocity vector.  b) What is the horizontal velocity?  c) What is the initial vertical velocity component? d) What time will the Beastie be in the air?  e) What distance will it go in that time?  f) what is the maximum distance you could make Adrock go with that speed?  (Use the range equation and plug in the angle which gives you the best range) g) Is Adrock the coolest Beastie Boy?
(14.9 m/s, 23 m/s, 4.7 sec, 70 m,76.6 m, no he's tied with Mike D and MCA)

8.

A projectile (a projectile is a physics term for what we would call a ball) leaves the ground with a speed of 34 m/s at an angle of 37o above the horizontal.  a) What is the initial velocity in vector component notation?  b) What time is the projectile (ball)  in the air? c) What is its range? d) What is its speed at the highest point? e) What is the velocity of the projectile (ball) in vector component notation when it is on the way up at elevation 10 m? f) Speed at elevation 10 m? g) Is the 5% we get from this assignment really worth all this work?
(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!)
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Solutions to Sample Problems

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1.

Brian's car, the Quantum, leaves the ground with an initial vertical velocity of 53 m/s and a horizontal velocity of 42 m/s.
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. 53² + 42² = initial speed².  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_xt  =  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.

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2.


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_xt = 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 3.3² + 11.76² = the hypotneuse² which we know as SPEED!  So the hypotenuse equals (square root)150 or 12.2!  Finding the height is easy.  Use y = 1/2gt² and voila u get 7.1 m.

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3.

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 gt² 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_xt = 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.
 

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4.

A Dustin is fired at an angle 35^o above the horizontal at a speed of 720 m/s.  a) Draw a picture of the initial velocity vector.  b) What is the horizontal velocity?  c) What is the initial vertical velocity component? d) What time will the shell be in the air?  e) What distance will it go in that time? f) what is your favorite color?

(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?
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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)

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_yot - ¹/₂ gt² 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.
 
 

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6.

Leo wants to jump into a big vat of frenchie cheese from a cliff that is 21 m tall.  There are rocks that stick out 3.4 m from the base of the cliff.  What time will he be in the air?  What must his speed be in order to clear the rocks? Why is he soooo frenchie?

(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_yot - ¹/₂ gt² 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) t²

This 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).

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7.

Adrock leaves the ground at a speed of 27.4 m/s at an angle above the horizontal of 57^o.  a) Draw a picture of the initial velocity vector.  b) What is the horizontal velocity?  c) What is the initial vertical velocity component? d) What time will the Beastie be in the air?  e) What distance will it go in that time?  f) what is the maximum distance you could make Adrock go with that speed?  (Use the range equation and plug in the angle which gives you the best range) g) Is Adrock the coolest Beastie Boy?

(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_yot - ¹/₂ gt²  and you get  0 = V_yot - ¹/₂ gt²

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_xt.  Multiplying the horizontal velocity by the expression we got earlier will eventually give you the equation x = (v₀² 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.
 
 

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8.

 A projectile (a projectile is a physics term for what we would call a ball) leaves the ground with a speed of 34 m/s at an angle of 37o above the horizontal.  a) What is the initial velocity in vector component notation?  b) What time is the projectile (ball)  in the air? c) What is its range? d) What is its speed at the highest point? e) What is the velocity of the projectile (ball) in vector component notation when it is on the way up at elevation 10 m? f) Speed at elevation 10 m? g) Is the 5% we get from this assignment really worth all this work?

(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_yot - ¹/₂ gt² 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!

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