Solving Differential Equations: Linear Models Homework

[Arshad_Physic]

Homework Statement

An Archer on top of a 50-ft high cliff shoots a 4-oz arrow straight up in the air. The bow is strung so that it can achieve maximum initial velocity of 128 feet per second. If the air resistance causes a drag constant of 1/1024 lb/fps, what is the maximum elevation reached by the arrow?

Homework Equations

mg-kv = m dv/dt

The Attempt at a Solution

So, here is the working I have done. But my answer is coming out to be WRONG. The answer should be 244 ft.

m=4 oz = 0.25 lb
k = 1/1024

mg-kv = m dv/dt

v(0) = 128
x(0) = 50

8 - v/1024 = 0.25 dv/dt

Multiply both sides by 1024 to make things easier:

8192 - v = 256 dv/dt

Separable equation, so after solving, I get:

v = Ae^(t/256) + 8192

Use the given condition: v(0) = 128 :

128 = A + 8192,
A = -8064

v(t) = -8064e^(t/256) + 8192

x(t) = integral[v(t)]

Thus, x(t) = -8064*256 e^(t/256) + 8192t + c
= -2064384e^(t/256) + 8192t + C

Using given condition x(0) = 50
c = 2064434

Thus, x(t) = -2064384e^(t/256) + 8192t + 2064434

on the equation: v(t) = -8064e^(t/256) + 8192, I use the condition v(? time) = 0

I get: t = 4.0316 seconds

I plug this in my x(t) equation and I get 307 feet.

This is WRONG! The answer should be 244 feet!

PLEASE HELP! :)

Thanks!

Arshad

 

[tiny-tim]

Hi Arshad!

(try using the X² tag just above the Reply box )

You keep getting + and - mixed up …

mg-kv = m dv/dt

Nooo …

8192 - v = 256 dv/dt

Separable equation, so after solving, I get:

v = Ae^(t/256) + 8192

Nooo …

 

[Arshad_Physic]

Ahh no - AFter integrating, the signs change and I get this answer. In one of the examples I have, the same thing happens to it as well.

When we integrate "8192 - v = 256 dv/dt", then we get:

-ln(8192-v) = t/256 + c

thus, it becomes v = Ae^(t/256) + 8192

PS: I should had used the x² button lol - sorry! :)

 

[tiny-tim]

-ln(8192-v) = t/256 + c

thus, it becomes v = Ae^(t/256) + 8192

No, e^-t/256 …

and what about your mg-kv = m dv/dt ?

 

[Arshad_Physic]

ohh! Now I see my mistake! But still the asnwer is not coming right! :O

okay, so:

v = Ae^-t/256 + 8192


128 = A + 8192,
A = -8064

v(t) = -8064e^(-t/256) + 8192

x(t) = integral[v(t)]

Thus, x(t) = -8064*-256 e(^t/256) + 8192t + c
= 2064384e(-t/256) + 8192t + C

Using given condition x(0) = 50
c = -2064434

Thus, x(t) = 2064384e^(-t/256) + 8192t - 2064434

on the equation: v(t) = -8064e(-t/256) + 8192, I use the condition v(? time) = 0

I get: t = -4.0316 seconds => (This is impossible, right?)

I plug this in my x(t) equation and I get -208.7 feet as my answer.

This is STILL WONG...The answer should be 244 ft...

This problem is giving me such a headache! Lol ! :)

 

[tiny-tim]

Are you still using mg-kv instead of -mg-kv ?

 

[Arshad_Physic]

Oh, I didn't understood that you were pointing that mistake of mine too! lol

But still, after using -mg-kv, I am getting t = 3.969 and my vertical distance is coming out to be 303.36 m...it should be 244 meters...

by the way, thanks a million for all the help you're giving! :)

 

[tiny-tim]

hmm …

perhaps the best thing is for you to sleep on it :zzz:, and write it all out again tomorrow …

maybe it'll be different then!

 

[Arshad_Physic]

lol

I will try putting the question to some other helping forums I guess! :) Or else post it here again - maybe someone else can figure it out! :)

Thanks a ton! :)

 

[Arshad_Physic]

...can someone else figure out the problem and help me?