Help with horizontal velocity

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    Horizontal Velocity
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Homework Help Overview

The problem involves a scenario where a rocket is launched from a height at an angle of 45 degrees, and the original poster seeks to determine another launch angle that would result in the rocket landing in the same spot, while ignoring air resistance. The subject area pertains to projectile motion and the effects of launch angles on horizontal range.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the significance of the 45-degree angle in relation to range, with one suggesting the use of trigonometric functions to find horizontal and vertical components of velocity. There is an emphasis on understanding the equations of motion and the geometry involved.

Discussion Status

The discussion is ongoing, with some participants providing guidance on the relevance of trigonometry and the properties of projectile motion. There is acknowledgment of the original poster's frustration and a request for more effort in problem-solving. Multiple interpretations of the problem are being explored, particularly regarding the use of trigonometry and conceptual understanding.

Contextual Notes

The original poster indicates a lack of familiarity with trigonometry and expresses concern about the approaching deadline for the assignment. There are hints of strict moderation in the forum, suggesting that participants are encouraged to demonstrate effort in their inquiries.

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Help pleasezz with horizontal velocity

Homework Statement


Ms. Rob launches a rocket from the school roof at an angle of 45 degrees and observes it's horizontal range. What other angle could she launch the rocket so that it lands in the same spot, ignoring air resistance?


Homework Equations


V=d/t
D=v/t
D=1/2gt^2
V=gt
A^2+b^2+c^2


The Attempt at a Solution



I just can't get it at all I'm getting really frustrated and it's due Friday! :cry::frown:
 
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What is special about the angle of 45deg when it comes to the range?

You can find the horizontal and vertical components of velocity from trigonometry.
recall: sin(45)=cos(45)=1/√2

You can work out the equations of motion by sketching velocity-time graphs for the components.

Please make more of an attempt at problems before posting - there is some very strict moderation going on here at the mo and you have to show some little effort beyond saying "I have no idea". You do have some ideas - you must have spent at least a year in physics class learning how to solve problems in general: do try to apply that knowledge.
 
Last edited:
Thanks

Thanks for the help I'm kinda still in Conceptual physics 1 and I don't know about trigonometry yet. Is their a simpler way? I just started science this year...thanks AGIAN
 
You can always draw the triangle and measure - but since this is conceptual physics, you have had a result in your notes about what to do. There's something special about the range at 45degs.

Note: If you know trigonometry anyway - do use it.
 

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