Math Question

[15357]

Out of "Aplicative Mathematics" Question #24:

A launches an object. The Coordinates of the launch site is (34,22,0,0) (x,y,altitude,time). The object is sighted 4 time units later at (54,40,16,4). Disregard air resistance. Assume that gravitational acceleration(9.8m/s^2) is equal regardless of altitude. Where will the object land? Use coordinates to answer. If needed, round to the first decimal. You may use a calculator if you wish.

a) When thrust (acceleration) is cut at point of launch:

b)When thrust is cut at 4 time units after launch:

c) When thrust is cut at 10 time units after launch:

d) What is the initial speed, if thrust is cut at point of launch?:

Bonus) What is the angle of launch? If needed, round to the first decimal.


I think I've got the question translated right. I think you're supposed to regard altitude as in meters (edit: Or kilometers. Try that.)... and time units as in seconds.


Have fun

 

[Dodo]

numbers, how do we know your not trying to caculate nuclear weapon platform maths huh?!!? : O
Anyway, i'd say 42

-dodo

 

[Raziaar]

This sounds like some of the projectile motion I've been learning from this physics dvd.

Except it's in 3D. And less known information :O

And well, I can't help you. Good luck!

 

[bbson john]

a) (58,44,0,4.8) (angle of elevation 74.1 degree, Projection angle N49E degree)

Am I incorrect?

 

[Solaris]

I think I could do that.

It's simple equations of motion. Xvuat stuff.

 

[15357]

numbers, how do we know your not trying to caculate nuclear weapon platform maths huh?!!? : O
Anyway, i'd say 42

-dodo

The actual question was about "Country A" launching "ballistics". But I decided to shorten that.

Hitchhiker?

a) (58,44,0,4.8) (angle of elevation 74.1 degree, Projection angle N49E degree)



Am I incorrect?

:/ If that's correct, I might have gotten the numbers wrong, I think it supposed to go much further. Or maybe altitude was in kilometers..... Damnit. I can't go check my textbook. Argh.

This sounds like some of the projectile motion I've been learning from this physics dvd.

Except it's in 3D. And less known information :O

And well, I can't help you. Good luck!

Oh well.

I think I could do that.

It's simple equations of motion. Xvuat stuff.

If you can, could you write down the way? 'Cause it's a bit hard for me to grasp all the questions. Especially the bonus.

 

[Solaris]

Okay; I'll do it now.

 

[Raziaar]

If you can, could you write down the way? 'Cause it's a bit hard for me to grasp all the questions. Especially the bonus.

I'm probably supremely wrong, but provided you find the answers to the other things, wouldn't you be able to find the angle of launch by creating a triangle out of the initial starting point, the top of the trajectory, and the landing point, and using the trigonometric function to find the angle using those vectors?

 

[bbson john]

b) (82,65.2,0,6.8) (angle of elevation 30.7 degree, Projection angle N48E degree)

Wrong again? :/

 

[15357]

Ty. Gotta go, though. See you tommorow

edit: bbson john, that is remarkable.

I'm probably supremely wrong, but provided you find the answers to the other things, wouldn't you be able to find the angle of launch by creating a triangle out of the initial starting point, the top of the trajectory, and the landing point, and using the trigonometric function to find the angle?

Probably, but then again, I hate trigonometry. So. :E

 

[99.vikram]

Can be done with vectors.

Can't be arsed to do it though. Maybe I'll post it sometime later.

 

[Solaris]

These two sites are all you need;

http://www.gcse.com/fm/motion.htm
http://www.antonine-education.co.uk/New_items/HFS/Jumping.htm

I started doing it but then got bored.

 

[Dan]

Out of "Aplicative Mathematics" Question #24:

A launches an object. The Coordinates of the launch site is (34,22,0,0) (x,y,altitude,time). The object is sighted 4 time units later at (54,40,16,4). Disregard air resistance. Assume that gravitational acceleration(9.8m/s^2) is equal regardless of altitude. Where will the object land? Use coordinates to answer. If needed, round to the first decimal. You may use a calculator if you wish.

I don't have a calculator, so these numbers will be approximate Numbers. I assume the coordinates are in metres, or that gravitational acceleration should be dimensionless. Sorry about my attempts at mathematical expressions. I wasn't very consistent, so there might be errors, but generally the answers worked out to convenient integers.

In summary
(58, 45, 0, 5)
(82,65,0,7)
(300,338,0,17)
25
26 degrees
(answers may or may not be correct)


Notations I used sorta inconsistently:
p=position/time coordinate
v=velocity
a=acceleration
d=displacement (I will indicate along which axis)
t=time

a) When thrust (acceleration) is cut at point of launch:

We start by looking only at altitude and time, x and y coordinates are non-accelerating so easy to find out given time and any two points. Over 4 seconds, altitude changes by +16, we know the acceleration, so it is easy to find the initial altitude.

Using calculus, or the equations of motion:

Solve for v0

t2 can now be solved from v0 and a. d2 is going to be 0 (back on the ground)
A really simple way is to just divide v0 into a, to find out how long it takes to reach the halfway mark, when v=0.

From here it is easy to go through the x and y coordinates, linearly extrapolating from the first 2 points.

Landing point is approximately (58, 45, 0, 5)

b)When thrust is cut at 4 time units after launch:

For this one, all axis are accelerating, but we can still treat x,y,z as independent problems related on the parameter of t because the axis are orthogonal. It must be assumed that the object is at rest at t=0 for any solution to exist.

Again, we start with altitude, because we need to find out how long it takes for altitude to reach 0. From p1 to p2, 16m of altitude is gained. This is under constant acceleration though. We already used displacement=v0t+1/2at^2 and v0 is 0.

16=1/2a*16
a=2

From here we can also find out that v1=8 (4 seconds of acceleration at 2m/s^2)

Now we just do the same as in the first part with v1 instead of v0 and solve for t
-16=8*t-4.9*t^2
-4.9t^2+8t+16=0
look familiar?

We want the positive root and a quick Google search on quadratic solvers tells us that t=2.8
So the object lands at t=4+2.8=6.8
p2=(x2,y2,0,6.8)

Now for x
Using the first equation in exactly the same way
a=(54-24)*2/16
a=5/2
v1=10 after 4 seconds

...and 2.8 seconds at that velocity will carry you 28m further.
so x2=82

Same thing for y
a=(40-22)*2/16
a=9/4
v1=9
and y2=65

So if I did everything right, you end up at.... (82,65,0,6.8)

c) When thrust is cut at 10 time units after launch:

This is kinda the same as b, but with an extra 6 seconds of acceleration tacked on.

First things first, altitude:
We already found out that the constant acceleration is 2. So 10 seconds of acceleration brings v1 up to 20. We also need to get displacement again at t=10. Just take the average velocity multiply time, and presto, altitude at t=10 is 100.

Now we do the same as the last question, solve for time when altitude is 0
-100=20*t-4.9*t^2
take the positive root again
t=7
So it takes 7 more seconds to hit the ground.

p2=(x2,y2,0,7)

Now solve for x axis
We know the acceleration is 5/2
After 10s, v=25, d=125
After 17s, d=300

Now solve for y axis
We know acceleration is 9/4
After 10s, v=22.5, d=112.5
After 17s, d=112.5+225=337.5

So p2=(300,337.5,0,17)

d) What is the initial speed, if thrust is cut at point of launch?:

Just add up the initial velocity vector magnitudes

vx=20/4=5
vy=18/4=4.5
vz=24

speed=sqrt(vx^2+vy^2+vz^2)
speed=sqrt(25+20+576)
speed=sqrt(621)
speed=25

Bonus) What is the angle of launch? If needed, round to the first decimal.

Well depends what that angle is in reference to. Angle above the horizon? I will assume that is what it is asking.
Take the tan of the vertical component, vz divided by the horizontal component, sqrt(vx^2+vy^2)
The answer is 26 degrees.

I think I've got the question translated right. I think you're supposed to regard altitude as in meters (edit: Or kilometers. Try that.)... and time units as in seconds.


Have fun

km will really mess up your answers because acceleration is still in m/s^2. Most rockets wouldn't gain 16 km altitude in 4 seconds.

 

[Raziaar]

Those equations are impossible to read in the blue style. You should save them out with a white background for people who use blue style. It's a little tedious copy and pasting them to open up in a new tab or window for the white background.

Doesn't matter to me though, since it's not my problem.

 

[Dan]

It's the way Texify writes them. I will do the rest in text because it is faster for me.

 

[Vegeta897]

thrust

That's what she said.

 

[Officer Combine]

That's what she said.

Your mother.