Solve Projetile Motion Homework: Max Height h = 0.25R tan theta

[xRadio]

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

 

[xRadio]

anyone?

 

[learningphysics]

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

Get another equation for maximum height...

 

[Doc Al]

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?

 

[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))

 

[xRadio]

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

 

[Doc Al]

0.5R?

That's a distance, not a time.

 

[xRadio]

That's a distance, not a time.

t= 0.5(R/vcostheta)?

 

[Doc Al]

t= 0.5(R/vcostheta)?

Good.

 

[xRadio]

so where does this take me? here?

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

 

[Doc Al]

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?

 

[xRadio]

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?

 

[Doc Al]

V1 = vsintheta?

That's the vertical component of the initial velocity.

 

[xRadio]

That's the vertical component of the initial velocity.

um no clue =(

 

[Doc Al]

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

 

[xRadio]

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

 

[Doc Al]

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

 

[xRadio]

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

Zero

 

[Doc Al]

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?

 

[xRadio]

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

 

[Doc Al]

vsintheta/2

Bingo! Now can you compute the max height?

 

[xRadio]

kk goott it thanks