Question

A mass of 0.2kg is released from rest. As the object falls, air provides a resistance proportional to the velocity (R(v) = 0.1v), where the velocity is measured in m/s. If the mass is dropped from a height of 50m, what is the velocity when it hits the ground?

i know that one way to solve it is to set up the equation using velocity as a function of time:

m * v' = -0.1v(t) - mg

then using this, find the position as a function of time, and THEN find where the time where the ball hits the ground.

I kinda know what to do but i'm having trouble

Answer 1

i'm thinking you solve the differential equation for v(t), then integrate it, THEN you can find what time the mass hits the ground

Answer 2

We know by Newton's second law \(\sum F=ma=m\dfrac{dv}{dt}\) and in our case we have two forces acting on our mass of \(,=0.2\), gravity \(-mg\) and air resistance \(0.1v\):$$0.2\frac{dv}{dt}=0.1v-0.2g\\2v'=v-2g\\\frac2{v-2g}dv=dt\\2\int\frac1{v-2g}dv=\int dt\\2\log(v-2g)=t+C\\\log(v-2g)^2=t+C\\(v-2g)^2=Ce^t\\v-2g=Ce^{t/2}\\v=Ce^{t/2}+2g$$We want \(v(0)=0\) so:$$0=Ce^0+2g\\C+2g=0\\C=-2g$$which means our velocity function is \(v=2g(1-e^{t/2})\). We want to determine the time we hit the ground so integrate:$$y=\int v\,dt=2g\int(1-e^{t/2})\ dt=2g(t-2e^{t/2})+C$$We wish for \(y(0)=50\) so solve:$$50=2g(-2e^0)+C=-4g+C\\C=50+4g$$So our equation for \(y\) is \(y=2g(t-2e^{t/2}+2)+50\). We wish to solve for \(t\) where \(y=0\):$$0=2g(t-2e^{t/2}+2)+50\\-50=2g(t-2e^{t/2}+2)\\-\frac{25}g-2=t-2e^{t/2}$$... which is difficult to solve. You need to use Lambert's W function.

Answer 3

thanks, I actually followed your work when i did it myself up to when you started integrated for the position function

Answer 4

I'm honestly lost here mate... do you know what the answer for the time should be?

Answer 5

a buddy of mine said he got around 28 m/s... but that answer is not definite

i do have access to the program Maple, so it will be able to use Lambert's W function if i can figure out how to work it

Answer 6

okay, i just plugged in my position equation into WolframAlpha and it solved t as -4.32s and 2.53s

so i plugged 2.53 into my velocity equation, and it gave me -49.84 m/s

Answer 7

Did you checkout the equation it tells you to in the problem set? It gives you a hint to look at (3.15):$$a=\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}=\frac{dv}{dy}v$$

Answer 8

thank you sir, i am set for now

Answer 9

http://www.math.uh.edu/~mflagg/ODE_HW1_copy_exercises.pdf
page 8 in the PDF, #12