Answer to Question #272534 in Differential Equations for Mizzy

Question

Consider that an object weighing 50 lb is dropped from a height of 1000ft with zero initial velocity. Assume that the air resistance is proportional to the velocity of the body. If the limiting velocity is known to be 200ft/sec, find the time it would take for an object to reach the ground.

Answer 1

From the second Newton's Law

$mv'=mg-kv$

Given $m=(50\times 0.453592)kg=22.6796kg$ and $g=32.1740ft/s^2$

$v'+{k\over 22.6796}v=32.1740$

$v'=-{k\over 22.6796}(v-{729.6934504\over k})$

${dv\over v-{729.6934504\over k}}=-{k\over 22.6796}dt$

We then integrate as follows

$v(t)={729.6934504\over k}-{729.6934504\over k}e^{-{kt\over 22.6796}}$

$v(t)\le200=\gt {729.6934504\over k}=200 \implies k=3.648467252$

$v(t)=200-200e^{-0.16087t}$

$v(t)=-h'(t)$

$h(t)=-\int vdt=-\int (200-200e^{-0.16087t})dt$

$=-200t-{200\over 0.16087}e^{-0.16087t}+c_2$

$h(0)=1000=-{200\over 0.16087}+c_2\implies c_2=2243.24$

$h(t)=-200t-1243.24e^{-0.16087t}+2243.24$

$h(t_1)=0=-200t_1-1243.24e^{-0.16087t_1}+2243.24$

And by solving this graphically, It will take 9.985 sec for the object to reach the ground