# Free Fall Differential Equations Problem - I am stuck

• February 14th 2010, 04:40 PM
Free Fall Differential Equations Problem - I am stuck
Question:

An Object of mass 5.00 kg is given an initial downward velocity of 74.5 m/sec and then allowed to fall under the influence of gravity. Assume the force due to air resistance (in N) on this object is twice its speed. If it hits the ground in exactly 10 seconds, how many meters above the ground was the object when it was released?

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The part I am stuck in this problem is what will be the drag co-efficient? It says "Assume the force due to air resistance (in N) on this object is twice its speed." - This is the part I DO NOT understand at all!!! :o

• February 15th 2010, 02:03 AM
chisigma
Setting $x(t)$ the quote of the onbject the equation describing the 'fall' is...

$m\cdot x^{''} = -m\cdot g +2\cdot x^{'} \rightarrow x^{''} -\frac {2}{m}\cdot x^{'}= -g$ (1)

The (1) is a linear second order constanr coefficients DE with initial conditions $x(0)=x_{0}$ [ $x_{0}$ is unknown] and $x^{'}(0)= -v_{0}$ and can be solved in standard way. After that the computation of $x_{0}$ is comfortable...

Kind regards

$\chi$ $\sigma$
• February 15th 2010, 02:51 AM
HallsofIvy
Quote:

Question:

An Object of mass 5.00 kg is given an initial downward velocity of 74.5 m/sec and then allowed to fall under the influence of gravity. Assume the force due to air resistance (in N) on this object is twice its speed. If it hits the ground in exactly 10 seconds, how many meters above the ground was the object when it was released?

---
The part I am stuck in this problem is what will be the drag co-efficient? It says "Assume the force due to air resistance (in N) on this object is twice its speed." - This is the part I DO NOT understand at all!!! :o

Are you serious? "Twice" means "multiply by 2"! The air resistance force is -2v. The total force, then, is -mg- 2v, due to gravity and air resistance.
$m\frac{dv}{dt}= -mg- 2v$ with v(0)= -74.5. You should be able to solve that equation for v(t). Integrating that will give x(t), the height at time t, with, of course, a constant of integration. Use the fact that x(10)= 0 to find that constant and then find x(0), the initial height.

Quote:

• February 15th 2010, 09:36 PM
lol! :) But thats what the confusion was; I had no idea that we will call it -2v lol - it now looks so simple after both of yours' help! :) lol

Ahh.........I wish that the authors would start writing GOOD books with EXAMPLES in them!! It will be costly lol, but atleast we will understand SOMETHING!!

lol!! THanks to BOTH of you - a LOT!!! :)