Projectile motion

[Aikenfan]

1. The problem statement, all variables and given/known data
http://i103.photobucket.com/albums/m121/aiken91919/Projectile.jpg


2. Relevant equations



3. The attempt at a solution

[Hootenanny]

At what angle are you launching the projectile? Are we taking air resistance into account?

[Aikenfan]

The problems given are all based on estimations...there was no angle given...air resistance can be left out, it really doesn't matter either way

[Hootenanny]

The problems given are all based on estimations...there was no angle given...air resistance can be left out, it really doesn't matter either way
Well, the most effective angle (requiring lowest launch velocity) would be \pi/4. So, tell me, have you any knowledge of kinematic equations?

[Aikenfan]

yes we have touched on them a little bit before

[Hootenanny]

Well, we can split the motion into two separate directions; horizontal motion and vertical motion. Can you use kinematic equations to describe the motion of the particle in each of these directions?

[Aikenfan]

http://i103.photobucket.com/albums/m121/aiken91919/U1L6a1.gif

[Aikenfan]

so the lowest launch velocity, would that be Vi

[Aikenfan]

So vi = pi/4?

[Saladsamurai]

V_i is the initial velocity. Pi/4 is the angle that requires the lowest magnitude of V_i. In oter words if theta_i is pi/4, you will get the most distance for any V_i. However, it is up to you to determine what V_i should be. I would personally start by converting the 9 miles into SI Units (I think metric would give some nice numbers to work with.

[fffff]

Yeah, one mile is equivalent to 1.68 km or something like that
or 1680m

So, since the horizontal component of velovity is contant

Range=vt (where v is horizontal component of velocity)

The vertical motion
vertical displacement=0

0=ut -5t^2 (where u is the vertical component of velocity)

[Aikenfan]

9 miles = 14.484096 kilometers

[Aikenfan]

i am going to correct that..it should be 8 miles...
8 miles = 12.874752 kilometers

[fffff]

yea, so the range= 12.9 km

on on related note, not enough data is given to solve the equation.
You need the time,t for the duration of projectile

[Saladsamurai]

but what are SI units?

[Aikenfan]

12 874.752 meters

[fffff]

shouldnt be too complicated, i uggest you make it t 3sf

[Aikenfan]

shouldnt be too complicated, i uggest you make it t 3sf what does t 3sf mean

[Saladsamurai]

Okay. Since you said that this problem can be based on estimates, let us call it 12.9 x10^3. (However for 9 miles I am getting 14.5x10^3...) Lokking back at your Projectile motion equations, which one do you think you should use?

[Aikenfan]

i realized it was a typo, im sorry, it should be 8 miles

[Saladsamurai]

which equation should you use?

[fffff]

3 significant figures.

[Aikenfan]

well, i only have distance, i am a little confused now...dont i need a time

[Saladsamurai]

well, i only have distance, i am a little confused now...dont i need a time
I believe that you should have one equation that does not involve time. Keep in mind that the paths of horizontal and vertical motion are independent of each other

Look at the equations again and pay special attention to the subscripts.
http://i12.photobucket.com/albums/a220/saladsamurai/u3l2e5.gif
http://i12.photobucket.com/albums/a220/saladsamurai/u3l2e4.gif

[Aikenfan]

Vf^2 = Vi^2 + 2 (a)(x)
Vf^2 = Vi^2 + 2 (a)(y)

do not involve time

[Saladsamurai]

You do not need time. ffff, all he needs to do is choose an angle to fire the projectile at and then calculate the initial Velocity it is to be fired at.

[Aikenfan]

12900 = 5t^2
= 50.79

[Aikenfan]

so i could choose pi/4 as the angle?

[Aikenfan]

would the distance need to be more than that because it is going up at an angle (i figured out the direct distance between the two schools)

[Saladsamurai]

I am sorry. I should have been more clear. You will need to use an equation (actually, 2 of them) that involve time. However, you will not need the time. This should be a hint. What do you know about the final velocity of the projectile? What do you know about the acceleration in the x direction? the y direction? do you know how to solve 2 equations with two unknowns?

[Aikenfan]

How did u get that?

[fffff]

Vertical Displacement when the particle hits the ground=0m
0=usinǾ(t) - (1/2)(9.81)(t^2) - equation 1

Its negative, because acceleration is negative throughout the motion

the other equation is range, or horizontal distance covered

Keep in mind, there is no acceleration in the x direction. Hence the horizontal component of velocity remains constant throughout the motion.

12900=ucosǾ x t - equatin2


so you have 4.9t=usinǾ

and 12900=ucosǾ x t

can you solve these equations and eliminate t.

remember Ǿ=45 degrees for maximum range

[fffff]

for the first equation i used s=ut +1/2at^2

and fr 2nd

velocity=distance x time

[Saladsamurai]

ffff. I am sorry. You are almost correct. 355m/s is the horizontal component of the initial velocity. I believe that the magnitude of the initial velocity would better suit the problem statement.

[Saladsamurai]

Aikenfan. Do you understand how we are arriving at these answers? Do you know the relationship between the magnitude of V and the x and y components of V?

[Saladsamurai]

I am getting 503m/s.

[Saladsamurai]

What was your final answer again?

[Aikenfan]

im still really lost, this problem is more complicated than i am used to...i am going to read through this again maybe i will understand

[fffff]

Im 100% right

4.9t=usinØ

and 12900=ucosØ x t

4.9 (12900/ucosØ) = usinØ

4.9(12900/(cos45xsin45))=u^2

u^2=12650
u=355 m/s

how could be incorrect?

[Aikenfan]

s=ut +1/2at^2
what does the s and the ut stand for

[Saladsamurai]

Im 100% right


u=355 m/s

how could be incorrect?

Like I said, "This is the horizontal component of Velocity"

[Saladsamurai]

s=ut +1/2at^2
what does the s and the ut stand for

I believe he is using s for displacement and u for speed. . . but without subscripts I cannot tell if he is refering to horizontal or vertical motion.

[fffff]

s=ut +1/2at^2

is the equation for displacement.

Before you attemp these questions, you must know each equation,
There are 4 equations which you showed earlier (in the image)

U=final velocity
Now, im splitting U into two parts.
Horiontal and vertical
horiontal=ucosØ
vertical=usinØ

remember that u^2 = (usin)^2 + (ucosØ)^2

[fffff]

what, ucosØ is the horizontal component of velocity ??
u=resultant velocity

[Saladsamurai]

355 is the horizontal. To answer the original question, I believe it would be more accurate to say that the projectile would need to be launched at 45 degrees to the horizontal with an initial velocity of 502 m/s.

[Aikenfan]

0=usinǾ(t) - (1/2)(9.81)(t^2) - equation 1

Why do you subtract these?

[Saladsamurai]

Akienfan. Here are the equations you started with. I used two of them (the ones we had enough given info for) and solved for the variable t. Now, by setting the other sides of the equations equal to each other, I was able to eliminate t and solve for only V_initial. ffff did the same thing, I believe, but using different notation...here. Let me know if there is something you don't understand. I know it can be very confusing when things do not look the way they did in class.:smile:
http://i12.photobucket.com/albums/a220/saladsamurai/proj.jpg= approx. 502 m/s

[fffff]

You used v= u +at
you set v as 0 m/s
but thats the time for max height, not when the particles hits the ground again

[fffff]

0=usinǾ (T) - (1/2)(g)(T^2)

usinǾ (T) = (1/2)(g)(T^2)

2usinǾ(t) / g = T

[fffff]

thats equals to

12900=ucosǾ (T)

T = S/ucosǾ

2usinǾ(t) / g = S/ucosǾ

[fffff]

u^2 = gs / 2sinǾcosǾ

[Saladsamurai]

Interesting. I don't know how I didn't catch that. My apologies. So the value that I used for time of y should have been doubled. This would make ffff absolutely correct. What was throwing me off is that oddly enough, the horizontal component of my original "502" answer is 355 as well! Poop.

[fffff]

Na, we all make mistakes all the time.
Let us know if your having any problems Aikenfan .

[Saladsamurai]

Hey ffff, we have (had before this) the same number of posts!

Aikenfan, do you hate us yet?:tongue2:

[Aikenfan]

haha! thanks ya'll, im working on it...i will let you know how i do :-)

[Aikenfan]

2usinǾ(t) / g = T
i am still a little confused about where the numbers come from that go into this

[Saladsamurai]

You need to know the relationship between the initial velocity (magnitude) and the y-component of initial velocity. Have you gone over these in class?

[Aikenfan]

No, We havent done anything like this

[Saladsamurai]

Have you studied Trig functions yet? Sine, Cosine, Tangent?

[Aikenfan]

No not really, at least not very much

[Aikenfan]

I just got it...i wanted to say thank you!

[Saladsamurai]

Excellent:smile: