Projectile Force and Velocity?

[Arthur Albert]

Hi guys, Please, someone can help me to solve some simple equations?
I Need to shoot a projectile with 20 grams at 15 meters and 15°.
So, I'm triyng to calculate the velocity and Force and what I will need to make this succesfull.

My Calcs:

*************************************************************************
Velocity= √(R*g)/sen θ
V= √(15m * 9,81) /sen15
V= 46,86 m/s
***
I= ΔQ or Qend - Qinitial. (Q = moviment quantity. Q= m.V)

I= (0,020kg * 46,86 m/s) - (0,020kg * 0 m/s)
I= 0,937 N.s

***
(I = Impulse // I= Force * Time)

0,937 N.s = Force * 0.1s
Force = 0,937 / 0,1 = 9,37 N.
**************************************************************************

This means I need to put just 9,37 Newtons on 0,1 seconds to shoot a 20 grams projectile at 15 meters and 46,86m/s? (at 15°)

Thank you for your attention and sorry for my bad english.

 

[nasu]

Isn't the range formula with a sin(2θ)?

 

[BvU]

Hello AA, welcome to PF !

You want to be a bit more clear in your mission statement:
1. is the 15 degrees wrt horizontal ?
2. Is the 15 m in a horizontal direction ?

Your first calc equates potential energy from height with kinetic energy: ##{1\over 2} m (v\sin\theta)^2 = mgh ## hence ## v = \sqrt{2 g h}/\sin\theta ##.
So like Nasu I think there is a factor 2 missing, only in a different place.
Indeed your 46.86 m/s will only get the object to a height of 7.5 m (that factor 2).
But by then it has already traveled 56 m and it hits the ground at 112 m. Is that the idea ?

 

[Arthur Albert]

I looked my notes but didn't has any "2" in the formula, my error.

The intention is throw the projectile 15m in a horizontal direction like that: http://imgur.com/rswHEH0 .

About the formula : √2gh / sin θ.
This will not just increase my velocity?
About the 15°, how can I calculate with 0° (parallel to the ground) ?

Thank you for the help Natsu and BvU!
Have a good day guys...

 

[nasu]

If it's about horizontal range, then you can look up the right formula here, for example: http://en.wikipedia.org/wiki/Range_of_a_projectile
If you launch from ground level, the second formula will apply.

If you want to have 0 degrees angle you need to launch from some height, don't you?

 

[Arthur Albert]

I see now, thank you @nasu and @BvU .
I will do the calculations.!