Projectile Motion Using Differential Equations

**jayman** · October 12th 2012, 08:53 PM

Hi everyone. It seems I've forgotten how to solve this one. I was wondering if someone could help me remember how.

Think projectile motion, straight up/down no resistance. I point out the part I have a problem with near the end. (this is all in accordance with a book on applied mathematics.)

d²x/dt²= -g initial conditions are dx/dt=v₀ x(0) = 0
x(t) = -1/2*g*t² + v₀ *t

The book makes says that I can solve for t to get
t=v₀/g - how?
and then realize the next equation without any explanation.
X_max= v₀²/2g - how?

I would be very grateful for help and direction on this one. Thanks.

**MarkFL** · October 12th 2012, 09:06 PM

Since we are given:

$x(t)=-\frac{1}{2}gt^2+v_0t$

then differentiating with respect to $t$ , we find:

$v(t)=-gt+v_0$

Equating this to zero (at the apex of the trajectory which is $x_{\text{max}}$ , the velocity will be zero), we find:

$-gt+v_0=0$

$t=\frac{v_0}{g}$

Now, using this value for $t$ , we find:

$x_{\text{max}}=x\left(\frac{v_0}{g} \right)=-\frac{1}{2}g\left(\frac{v_0}{g} \right)^2+v_0\left(\frac{v_0}{g} \right)=$

$-\frac{v_0^2}{2g}+\frac{v_0^2}{g}=\frac{v_0^2}{2g}$

**jayman** · October 12th 2012, 09:30 PM

"A pattern so grand and complex..."

Can you explain v(t) from x(t)

Thanks Mark.... BTW I'm a HUGE Rush fan.

**MarkFL** · October 12th 2012, 09:39 PM

By definition, we have:

$v(t)\equiv\frac{d}{dt}x(t)$

This simply means that velocity is the time rate of change of displacement, or position.

So, if you have the displacement, you may differentiate it to get the velocity.

I was saddened a bit the other day to hear "Fly By Night" used in a commercial to sell Volkswagens.

**jayman** · October 12th 2012, 09:49 PM

LOL... to all of your last post... I must have missed that because it is late/early here... Can't believe I didn't see Calc I.

Yeah, that commercial is a bit... off.

Thank you again Mark. I can get back to studying in the morning. Good to know that there are great people on this sight with great answers.

**MarkFL** · October 12th 2012, 09:55 PM

By the way, I was raised in Evansville...I really miss the Fall Festival there!

**jayman** · October 13th 2012, 07:03 AM

The colors are wonderful this year.

**jayman** · October 13th 2012, 07:06 AM

BTW, how are you generating the equation graphics? They look great.

**MarkFL** · October 13th 2012, 09:01 AM

Those are generated using $\LaTeX$ . Do a search here and online for LaTeX usage, and you will soon be making nice looking expressions too!

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