Projetile Motion

1. Sep 13, 2007

1. The problem statement, all variables and given/known data
A soccer ball is kicked from the ground at an angle theta above the horizontal. Show that the equation h = 0.25R tan theta represents the maximum height of the ball, where h is the height and R is the range.

2. Relevant equations

t= dx/V1x
dy = v1y(t) + .5a(t)^2

3. The attempt at a solution

Vert - dy = 0, V1 = vsintheta, a = g, t=?
Horizontal - dx = R, V1 = vcostheta, t=R/vcostheta

dy = v1y(t) + .5a(t)^2
0 = vsintheta(R)/vCostheta + 1/2g(R/vcostheta)^2
sintheta(R)/Costheta = -1/2g(R/vcostheta)^2

Help plz

2. Sep 13, 2007

anyone?

3. Sep 13, 2007

learningphysics

Looks good... except you should have used -(1/2)gt^2...

Get another equation for maximum height...

4. Sep 13, 2007

Staff: Mentor

Realize that that's the time for the complete trajectory. How long does it take to reach the maximum height?

5. Sep 13, 2007

learningphysics

One simplification I wanted to point out:

0 = vsin(theta)t - (1/2)gt^2 (where t is the time to reach R)

can be simplified to

0 = vsin(theta) - (1/2)gt, where t = R/(vcos(theta))

6. Sep 13, 2007

0.5R?

I am completely loss and dont know where to go from the step i took it up to

Last edited: Sep 13, 2007
7. Sep 13, 2007

Staff: Mentor

That's a distance, not a time.

8. Sep 13, 2007

t= 0.5(R/vcostheta)?

9. Sep 13, 2007

Staff: Mentor

Good.

10. Sep 13, 2007

so where does this take me? here?

0 = sin(theta) - (1/2)(0.5(R/costheta))
(1/2)(0.5(R/costheta)) = sin(theta)

Last edited: Sep 13, 2007
11. Sep 13, 2007

Staff: Mentor

No.

Keep it simple. You know how long it takes to reach maximum height. What's the average (vertical) speed as it rises?

12. Sep 13, 2007

V1 = vsintheta?

13. Sep 13, 2007

Staff: Mentor

That's the vertical component of the initial velocity.

14. Sep 13, 2007

um no clue =(

15. Sep 13, 2007

Staff: Mentor

What's the vertical component of the velocity when it reaches maximum height?

16. Sep 13, 2007

(2d /0.5(R/vcostheta)) - vsintheta = V2?

17. Sep 13, 2007

Staff: Mentor

What if you threw a ball straight up? What would its speed be at its highest point?

18. Sep 13, 2007

Zero

19. Sep 13, 2007

Staff: Mentor

Yes! Now apply that fact to your problem. You know the initial (vertical) speed and the final speed: so what's the average (vertical) speed as the projectile rises?

20. Sep 13, 2007