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

AI Thread Summary
To determine the angle that yields the maximum horizontal range for a package dropped from a supply plane flying at 60.0 m/s and 200.0 m altitude, one must derive an expression for the range as a function of the angle θ. The horizontal motion has no acceleration, while vertical motion is influenced by gravity. The next steps involve calculating the time it takes for the package to hit the ground and then finding the corresponding horizontal distance at that time. To maximize the range, one can either differentiate the range function with respect to θ or complete the square. Understanding these concepts is crucial for solving the problem effectively.
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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.
 
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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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