(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

mg-kv = m dv/dt

3. 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

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