Calculating the Angle of a Projectile Fired for 3x Range Horizontally

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Homework Help Overview

The problem involves a projectile fired such that its horizontal range is three times its maximum height. Participants are exploring the relationship between the angle of projection and these dimensions in the context of projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use a geometric approach with arctan(1/3) to find the angle but expresses confusion about its validity. Other participants inquire about the relevant equations of motion and discuss the implications of the 1:3 ratio on the calculations.

Discussion Status

The discussion is ongoing, with participants questioning the setup and exploring different aspects of projectile motion. Some guidance has been offered regarding the equations of motion, but there is no clear consensus on the next steps or the correct approach to take.

Contextual Notes

There appears to be uncertainty regarding the correct equations to use and how to incorporate the 1:3 ratio into the analysis. One participant notes a potential error in the equation for vertical motion.

jesuslovesu
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A projectile is fired so that its horizontal range is 3 times its max height. What is the angle?

Well, I'm stumped because I thought if I just drew up a triangle and did
arctan(1/3), I would get the correct angle... apparently not.
 
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What are the equations of motion?
 
hmm

x = vx*t
y = -1/2g*t^2

I'm not really sure where to go, how the 1:3 would factor into anything.
 
jesuslovesu said:
hmm

x = vx*t
y = -1/2g*t^2

I'm not really sure where to go, how the 1:3 would factor into anything.
Your equation for y is incorrect.

Set up Newton's second law of motion for the system, both directions!
 

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