Solving Projectile Problem for Physics Assignment

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Homework Help Overview

The original poster discusses a physics assignment involving projectile motion, specifically determining the two launch angles that allow a projectile to reach a distance of 560 meters when launched at a velocity of 82 m/s, with takeoff and landing at the same height.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants suggest breaking down the motion into x and y components, while others propose using the range equation to simplify the problem. There is discussion about the relevance of different equations and methods for solving the problem.

Discussion Status

Some participants have offered guidance on using the range equation, while others emphasize the importance of understanding the derivation of equations rather than memorizing them. Multiple approaches are being explored without a clear consensus on the best method.

Contextual Notes

The original poster expresses frustration with the problem and notes a deadline for the assignment, indicating a sense of urgency. There is also mention of the assumption that the projectile's initial and final heights are the same.

garboui
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i have been give an asignment to do in physics. well everything is fine and dandyuntill the last question.

what the question wants to know is that for a projectile launched at a certain velocity(82m/s) hast to reach a set distance(560m), find the two angles that will allow the target distance to be reached. takeoff and landing are at the same height.

is there some formula where i could plug all of the numbers in and get the answer or is there something that i am missing here. i know its easy but for some reason i just can't figure it out for the life of me.

any help will be greatly appreciated as the assignment is due tommrow morning.

thanks
garboui
 
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Divide the motion into x and y-components, and deal with them as if you were dealing with one dimensional motion.

The equations:
[tex]\left\{ \begin{array}{l}<br /> x = x_0+v_{x0}t+\frac{1}{2}a_xt^2 \\ <br /> y = y_0+v_{y0}t+\frac{1}{2}a_yt^2 \\ <br /> \end{array} \right.[/tex]
 
Dont worry about those position equations. Though they work elegantly and efficiently, there is another, more efficient method. since all you have is the RANGE and the initial velocity, plus the fct that the initial and final height are the SAME, the range equation is your new best friend. By eliminating the time parameter and substitution, you get the more useful eqution: R=V0^2*sin(2theita)/g. the beauty here is that g is a given, along with Vo and R. just solve for theita and rember that there will be two angles, one slightly above and below 45 degrees.
 
DaMastaofFisix said:
Dont worry about those position equations. Though they work elegantly and efficiently, there is another, more efficient method. since all you have is the RANGE and the initial velocity, plus the fct that the initial and final height are the SAME, the range equation is your new best friend. By eliminating the time parameter and substitution, you get the more useful eqution: R=V0^2*sin(2theita)/g. the beauty here is that g is a given, along with Vo and R. just solve for theita and rember that there will be two angles, one slightly above and below 45 degrees.
True, that is the equation we get for this problem from the position equations. One should not in any mathematical subject learn equations by heart. Instead, one should learn how to derive them.

The position equations give far more information on the projectile's movement than the "range equation".
 

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