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

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a soccer ball kicked at an angle theta. Participants are tasked with demonstrating 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.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between time, range, and maximum height, questioning how to derive the maximum height from the given parameters. There are discussions about the equations of motion and the time taken to reach maximum height.

Discussion Status

Some participants have provided guidance on simplifying equations and understanding the time to reach maximum height. There is an ongoing exploration of the average vertical speed during the ascent, with various interpretations of the equations being discussed.

Contextual Notes

Participants express confusion regarding the distinction between distance and time, and there are references to specific values and relationships that are not fully resolved. The discussion reflects the complexity of the problem and the need for further clarification on certain assumptions.

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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
 
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anyone?
 
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
 
Last edited:
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.
 
  • #10
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:
  • #11
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?
 
  • #12
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?
 
  • #13
xRadio said:
V1 = vsintheta?
That's the vertical component of the initial velocity.
 
  • #14
Doc Al said:
That's the vertical component of the initial velocity.

um no clue =(
 
  • #15
What's the vertical component of the velocity when it reaches maximum height?
 
  • #16
(2d /0.5(R/vcostheta)) - vsintheta = V2?
 
  • #17
What if you threw a ball straight up? What would its speed be at its highest point?
 
  • #18
Doc Al said:
What if you threw a ball straight up? What would its speed be at its highest point?

Zero
 
  • #19
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?
 
  • #20
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
 
  • #21
xRadio said:
vsintheta/2
Bingo! Now can you compute the max height?
 
  • #22
kk goott it thanks
 
Last edited:

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