Proving maximum height h = 0.25R tan θ for projectile motion

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Homework Statement


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.


Homework Equations



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

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 please
 

Discussion

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

Get another equation for maximum height...
 
xRadio said:
Horizontal - dx = R, V1 = vcostheta, t=R/vcostheta
Realize that that's the time for the complete trajectory. How long does it take to reach the maximum height?
 
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))
 
Doc Al said:
Realize that that's the time for the complete trajectory. How long does it take to reach the maximum height?
0.5R?

I am completely loss and don't know where to go from the step i took it up to
 
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xRadio said:
0.5R?
That's a distance, not a time. :wink:
 
Doc Al said:
That's a distance, not a time. :wink:

t= 0.5(R/vcostheta)?
 
xRadio said:
t= 0.5(R/vcostheta)?
Good.
 
so where does this take me? here?

0 = sin(theta) - (1/2)(0.5(R/costheta))
(1/2)(0.5(R/costheta)) = sin(theta)
 
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xRadio said:
so where does this take me? here?

0 = sin(theta) - (1/2)(0.5(R/costheta))
No.

Keep it simple. You know how long it takes to reach maximum height. What's the average (vertical) speed as it rises?
 
Doc Al said:
No.

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

V1 = vsintheta?
 
xRadio said:
V1 = vsintheta?
That's the vertical component of the initial velocity.
 
Doc Al said:
That's the vertical component of the initial velocity.

um no clue =(
 
What's the vertical component of the velocity when it reaches maximum height?
 
(2d /0.5(R/vcostheta)) - vsintheta = V2?
 
What if you threw a ball straight up? What would its speed be at its highest point?
 
Doc Al said:
What if you threw a ball straight up? What would its speed be at its highest point?

Zero
 
xRadio said:
Zero
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?
 
Doc Al said:
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?

vsintheta/2
 
xRadio said:
vsintheta/2
Bingo! Now can you compute the max height?
 
kk goott it thanks
 
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