Projectile motion

1. May 13, 2007

Aikenfan

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 13, 2007

Hootenanny

Staff Emeritus
At what angle are you launching the projectile? Are we taking air resistance into account?

3. May 13, 2007

Aikenfan

The problems given are all based on estimations...there was no angle given...air resistance can be left out, it really doesn't matter either way

4. May 13, 2007

Hootenanny

Staff Emeritus
Well, the most effective angle (requiring lowest launch velocity) would be $\pi/4$. So, tell me, have you any knowledge of kinematic equations?

5. May 13, 2007

Aikenfan

yes we have touched on them a little bit before

6. May 15, 2007

Hootenanny

Staff Emeritus
Well, we can split the motion into two separate directions; horizontal motion and vertical motion. Can you use kinematic equations to describe the motion of the particle in each of these directions?

7. May 16, 2007

Aikenfan

8. May 16, 2007

Aikenfan

so the lowest launch velocity, would that be Vi

9. May 16, 2007

Aikenfan

So vi = pi/4?

10. May 16, 2007

Saladsamurai

V_i is the initial velocity. Pi/4 is the angle that requires the lowest magnitude of V_i. In oter words if theta_i is pi/4, you will get the most distance for any V_i. However, it is up to you to determine what V_i should be. I would personally start by converting the 9 miles into SI Units (I think metric would give some nice numbers to work with.

11. May 16, 2007

fffff

Yeah, one mile is equivalent to 1.68 km or something like that
or 1680m

So, since the horizontal component of velovity is contant

Range=vt (where v is horizontal component of velocity)

The vertical motion
vertical displacement=0

0=ut -5t^2 (where u is the vertical component of velocity)

12. May 16, 2007

Aikenfan

9 miles = 14.484096 kilometers

13. May 16, 2007

Aikenfan

i am going to correct that..it should be 8 miles...
8 miles = 12.874752 kilometers

14. May 16, 2007

fffff

yea, so the range= 12.9 km

on on related note, not enough data is given to solve the equation.
You need the time,t for the duration of projectile

15. May 16, 2007

Saladsamurai

but what are SI units?

16. May 16, 2007

Aikenfan

12 874.752 meters

17. May 16, 2007

fffff

shouldnt be too complicated, i uggest you make it t 3sf

18. May 16, 2007

Aikenfan

what does t 3sf mean

19. May 16, 2007

Saladsamurai

Okay. Since you said that this problem can be based on estimates, let us call it 12.9 x10^3. (However for 9 miles I am getting 14.5x10^3...) Lokking back at your Projectile motion equations, which one do you think you should use?

20. May 16, 2007

Aikenfan

i realized it was a typo, im sorry, it should be 8 miles

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