# Homework Help: Very simple yet hard for me

1. Sep 1, 2011

### gheelengooi

Dear All,

I am figuring out the initial velocity of a projectile in order to do a estimation.

I have a horizontal force of F pushing a projectile horizontally (meaning a horizontal launch) from a height H. I assume there is no more force pushing it afterwards.

I want to know how far horizontally that thing would travel. I have the height of the projectile to the ground so what I would need is actually a simple step of converting that initial force into initial velocity. And yes I have the mass of the projectile.

Thank you so much for the help!

GLO

2. Sep 1, 2011

### Staff: Mentor

You'll have to supply more information. As you realize, what's needed is the object's speed when launched.

How many forces act? For how long? If the horizontal force is the only force on the object and it is constant, then you can figure out the acceleration and use kinematics to find the speed.

3. Sep 1, 2011

### gheelengooi

Thank you for your reply. There is only one horizontal force acting. It is the only force. You can't say it's constant cause it is applied only once, which is on the projectile itself. Afterwards the only force acting would be the gravity.

Thank you very much!

4. Sep 1, 2011

### daveb

You'll also need the time which the horizontal force is applied. A 100N force acting for 1 millisecond produces a much different intial velocity than a 100N force acting for 1 microsecond.

5. Sep 1, 2011

### Staff: Mentor

That's rather mysterious. Do you apply that force for some time or over some distance? To really compute the speed you'll need the force applied as a function of time.

Why don't you describe exactly what it is that you are doing. How are you applying this force to the projectile?

6. Sep 1, 2011

### gheelengooi

Dear All,

Sorry for being mysterious. I thought the info would be sufficient.

Okay here's what I am doing. I wanted to do a sprinkler system for my experiment setup, and I plan to use the FogCo nozzles which would atomize the water particles into mist, so that it would provide water to the soil yet will not create erosion.

The nozzle itself operates at a minimum 250 psi, and currently in my lab I do not have the setup which operates at such a high pressure. So the catch is, I have to design the thing first only I get to test it, which is pretty stupid but I am forced to do that anyway.

In order to provide a good estimation to the area of water which the nozzle could spray out, I decide to use very simple physics:-

1) From the nozzle orifice which is 0.508 mm, use the pipe to orifice equation and a constant c of 0.98 since it is a rounded outlet to calculate the head drop across the orifice.
2) Assume no friction loss (yet, but can always add later), use 250 psi minus the head drop and that yields the net pressure acting on the water before they got sprayed out from the nozzle orifice.
3) Since the atomisation of the water droplet is up to 5 to 15 microns, I take 10 microns as my calculation reference. Assume water droplet is spherical at such small scale and thus air friction is negligible. From the size of water droplet (take diameter = 10 micron), you can estimate the volume of one water droplet and thus its mass.
4) Assume the edge of the cone shape water area sprayed by nozzle is actually those water droplets pressurized out from the orifice at a 180 deg (totally horizontal) angle, thus it is possible to calculate how far the water droplet would drop.

I have the minimum height of the nozzle to the soil, so that's why H is already known. You can assume that since the water droplets are perfectly spherical, thus the force applied on the droplet once they escape from the orifice is simply P/A, where A = pi*r^2, and r = 5 micron.

I hope that clarifies how I come to this part and I just had to figure out how to account for the initial velocity of the little water droplet.

Thank you all again!!

Best,
GLO

7. Sep 2, 2011

### gheelengooi

Dear All,

Just wanna push this post again so that if anyone has any idea please do help!

Thank you so much!

Best,
GLO