projectile problem

[neil]

Hi, I was wondering if anyone can help with a problem that has me stumped.

I know the starting position & the landing position of a projectile. (on a level surface)
I also know the initial angle the projectile will be fired at.

I need to find the initial launching velocity of the projectile.
And then how would this be adjusted to take air resistance into account?

Thanks for any help.

N.

[marcus]

Originally posted by neil
Hi, I was wondering if anyone can help with a problem that has me stumped.

I know the starting position & the landing position of a projectile. (on a level surface)
I also know the initial angle the projectile will be fired at.

I need to find the initial launching velocity of the projectile.
And then how would this be adjusted to take air resistance into account?

Thanks for any help.

N.

I see the problem's been read over a dozen times by others and not replied to.

I'll help a little way along---just the part with no air resistance.

v2 = gD/2sinAcosA

where g is gravity
D is the range---horiz dist traveled
A is the angle of elevation

You may have gotten that already and be waiting for help with
air resistance (which you wont get from me)

In all events, you should certainly understand this case very well before proceeding.

Do you understand the air-less case?

1. vert vel = vsinA
2. g T/2 = vsinA
3. T = 2vsinA/g
4. horz vel = vcosA
5. D = T vcosA
6. D = (2vsinA/g)vcosA
7. ...... next step?

[enigma]

There is no simple equation for a projectile with air resistance.

Air resistance changes with velocity, and its direction is also changing as the projectile travels along its path.

The only way to do it is to update the direction and magnitude of the drag in steps. If you're still learning the air-less case, going through the procedure will probably be more trouble than its worth.

What do you need it for?

[Alexander]

Originally posted by enigma
There is no simple equation for a projectile with air resistance.



Except for a one-dimensional motion, like a fall directly down. Then there is reasonably simple equation with analytical solution (something like v(t)=voth(gt/vo), where vo is terminal speed).

[schwarzchildradius]

If you need it very bad, I have a worked out solution for this problem from Symon's Mechanics, but in principle you just have to write down Newton's law of motion:
md2x/dt2 = F
where F = mg -bv {in the case of friction}
and solve for x, which is not fun.

[Alexander]

This one can be written for velocity (v=x') as mv'+bv-mg=0 which results in v(t)=vo(exp(-bt/m)+vterm(1-exp(-bt/m)), where vo is initial velocity and vterm=mg/b (usually named as "terminal" velocity).

Position x(t) is just an integral of this exponent: x(t)= m(vterm-vo)exp(-bt/m)/b +vtermt+xo