Skydiver

Calculus Level 1

Two forces act on a parachutist. One is $mg,$ the attraction by the earth, where $m$ is the mass of the person plus equipment and $g=9.8 \text{ m/sec}^2$ is the acceleration of gravity. The other force is the air resistance ("drag"), which is assumed to be proportional to the square of the velocity $v(t)$ .

Using Newton's second law of motion (mass $\times$ acceleration = net force applied), set up an ordinary differential equation for $v(t).$

Let $k$ denote the drag coefficient.

mg-k{v}^{2}=m\frac { d{v}^{2} }{ d{t}^{2} }

mg-{k}^{2}{v}^{2}=m\frac { dv }{ dt }

mg-k{v}^{2}=m\frac { dv }{ dt }

mg=m\frac { dv }{ dt }

2 solutions

Jeel Shah
Jul 7, 2016

We are given the following information: $mg$ , $k$ and $v$ and we are asked set up an ODE for $v(t)$ using Newton's Second Law which says that $F_{net} = ma$ .

Therefore, we have to figure: 1. How to set up an equation for $v(t)$ 2. Determine which forces are applied on the parachutist.

(1). We know that $a$ in $F_{net} = ma$ is really just rate of change of velocity or $\frac{dv}{dt}$ so we can transform law into

$F_{net} = m\frac{dv}{dt}$

(2). The first paragraph explains the forces which are being applied i.e. $mg$ (our given) and "the air resistance (drag), which is assumed to be proportional to the square of the velocity $v(t)$ . Furthermore, we are told that drag is represented by $k$ . So putting these pieces together we see that the total force on the parachutist is $mg - kv^2$ .

So finally we have the answer:

$\boxed{mg - kv^2 = m\frac{dv}{dt}}$

Thanks for the solution! It is indeed very well explained. However, we do need to state that we are considering the downward direction as "positive" or else the equation could be different :)

Raivat Shah - 3 years ago

Joe Potillor
Dec 5, 2016

1) what is the limiting velocity? 2) would the parachute still be sufficient if the air resistance was only proportional to v(t)?

shubh shah - 3 years, 3 months ago

Skydiver

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