Proving Projectile Farthest Flight at 45° Angle

  • Context: High School 
  • Thread starter Thread starter LENIN
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary

Discussion Overview

The discussion centers around the question of proving that a projectile achieves the farthest flight when launched at a 45-degree angle. It involves exploring the equations of motion and the conditions under which this angle maximizes horizontal distance, particularly in the context of ideal projectile motion without air resistance.

Discussion Character

  • Exploratory, Technical explanation, Homework-related, Mathematical reasoning

Main Points Raised

  • One participant asserts that a projectile flies the farthest when launched at a 45-degree angle but seeks assistance in proving this claim.
  • Another participant requests the equations of motion related to projectile motion.
  • A participant clarifies that the 45-degree angle assumption holds true only in the absence of air resistance and provides the equations for horizontal and vertical components of motion.
  • This participant outlines a method to find the time of impact and the horizontal distance traveled, suggesting differentiation to find the angle that maximizes distance.
  • A later reply expresses gratitude for the guidance and mentions progress in understanding the steps, particularly the use of differentiation, despite being new to the topic.

Areas of Agreement / Disagreement

Participants generally agree on the premise that the 45-degree angle is optimal under certain conditions, specifically the absence of air resistance. However, the discussion remains open regarding the proof and the specific mathematical steps involved.

Contextual Notes

The discussion does not address the effects of air resistance or other real-world factors that may influence projectile motion. There are also unresolved mathematical steps related to maximizing the horizontal distance.

Who May Find This Useful

This discussion may be useful for students learning about projectile motion, particularly those interested in the mathematical derivation of optimal launch angles and the underlying physics principles.

LENIN
Messages
101
Reaction score
1
I know that a projectile flyes the farthest, when it is louncht under a 45 degree angel. But I don't know how to proofe this. Any help would be apreeciated.
 
Physics news on Phys.org
Well,then can u write down the equations of motion...?

Daniel.
 
45 degrees assuming no air resistance!

If the angle is θ degrees, initial speed v, then the horizontal and vertical components of initial velocity are vcos(θ) and vsin(&theta) respectively.

acceleration is 0 horizontally, - g vertically so
velocity (t)= vcos(θ) horizontally, vsin(θ)- gt vertically.

position(t)= vcos(θ)t horizontally, vsin(θ)t- (1/2)gt2 vertically. (taking (0,0) as starting point.)

1. Solve for the time of impact (i.e. set height= 0, find non-zero t solution- it will depend upon θ).

2. Find horizontal distance at that time (again, it will depend upon θ).

3. Find the value of θ that maximizes that (differentiate with respect to θ, set equal to 0).
 
Thenks! I got to the 2'nd step myself but the third one is really helpful. I didn't think about useing differentilas, becouse I only started takeing them a few weaks ago and I still don't find my way around very well. Thenks agein.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 60 ·
3
Replies
60
Views
8K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K