(2D problem) What angle will yield maximum horizontal range?

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Homework Help Overview

The problem involves determining the angle of flight that will yield the maximum horizontal range for a package dropped from a supply plane flying at a specific speed and altitude. The subject area pertains to projectile motion in two dimensions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations related to projectile motion and question the assumptions regarding acceleration in the x-direction. There are attempts to derive an expression for the range as a function of the angle and to explore methods for finding the maximum range.

Discussion Status

The discussion is ongoing, with participants providing guidance on deriving the range expression and exploring methods for maximizing it. There is a mix of uncertainty and attempts to clarify the next steps in the problem-solving process.

Contextual Notes

Participants are navigating the complexities of combining equations for horizontal and vertical motion, and there is mention of using numerical methods to test angles, indicating a potential lack of clarity in analytical approaches.

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Homework Statement



A supply plane is flying at 60.0m/s at an altitude of 200.0m. What angle of flight will yield the maximum horizontal range for the dropped package?

Homework Equations


1/2(g)(t)2.


The Attempt at a Solution


So far the only equations that i have, but may be wrong, is x=((60m/s)cos(\Theta))t+1/2(g)(t)2. and y=((60m/s)sin(\Theta))t+1/2(g)(t)2+200m. I'm told it would involve a quadratic formula if i do it the more efficient way, but i cannot figure out how to complete it, or if i am going about it all wrong to find the angle that gives the farthest distance. I have also tried messing around with the formulas to try to combine them. I've also thought of using excel to test every angle from 0.0-90.0° but i haven't worked out how i would do that either.
 
Last edited:
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Welcome to PF!

Hi Wily Fury! Welcome to PF! :smile:
Wily Fury said:
x=((60m/s)cos(\Theta))t+1/2(g)(t)2

nooo :redface:

in the x direction, a = … ? :wink:
 
the acceleration should be 0 for the x direction?
 
Yup! :biggrin:
 
i'm still not sure how to approach the next step or if i am on the right track.
 
Wily Fury said:
i'm still not sure how to approach the next step or if i am on the right track.

The next step is to get an expression for the range (as a function of θ).

Then find the maximum value of the range, either by differentiating wrt θ, or by completing the square. :wink:

(and now I'm off to bed :zzz:)
 
I'm not really sure what that means >.<
 
what what means? :confused:
 
Wily Fury said:
i'm still not sure how to approach the next step or if i am on the right track.
You have (correcting your sign for vertical acceleration) x = u cos(θ)t, y = h + u sin(θ)t - gt2/2. Try to answer the following in sequence:
- at what time (as a function of h, u, g, theta) will the package hit the ground?
- what will x be at that time?
- how do you maximise that wrt theta?
 

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