Since we are given:
then differentiating with respect to , we find:
Equating this to zero (at the apex of the trajectory which is , the velocity will be zero), we find:
Now, using this value for , we find:
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.)
d^{2}x/dt^{2}= -g initial conditions are dx/dt=v_{0} x(0) = 0
x(t) = -1/2*g*t^{2} + v_{0} *t
The book makes says that I can solve for t to get
t=v_{0}/g - how?
and then realize the next equation without any explanation.
X_{max}= v_{0}^{2}/2g - how?
I would be very grateful for help and direction on this one. Thanks.
Since we are given:
then differentiating with respect to , we find:
Equating this to zero (at the apex of the trajectory which is , the velocity will be zero), we find:
Now, using this value for , we find:
By definition, we have:
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.
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.