Projectile Motion with Air Resistance problem

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The discussion revolves around determining the optimal launch angle for a projectile with air resistance to achieve maximum range. The original poster suggests that the angle should be 53 degrees, as indicated by their teacher, and seeks proof for this assertion. Participants note that the equations governing projectile motion with air resistance are complex, often requiring numerical solutions due to the nature of drag forces. It is generally understood that, without air resistance, the optimal angle is 45 degrees, and many contributors argue that the optimal angle with air resistance is typically less than 45 degrees. The conversation highlights the need for clarification on the specific air resistance formula used in the calculations.
ahadpoolghaha
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hi.
Suppose that a projectile of mass "m" is launched, from ground level in a horizontal plain, making an angle teta to the horizontal .(with air resistance).now in which angle "m" has the maximum range? .(our teacher said that we should find 53 degrees for teta).please prove it.
(first of all we should find (x-t) and (v-t) equations.then we should find the equation of range with those 2 equations and then we should derivative the rang equation(dr/d(teta).
so we will find the angle in which the mass "m" has the maximum range.this is my idia.if anyone knows the answer of this question,please tell it.i need the diagram of the motion too.)thanks a lot.:smile:
 
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I think this thread should be in the homework location... none the less you state with air resistance. My understanding is that the equation that describes air resistance is pretty complicated and is normally simplified to either linear or quadratic approximation. But Newtons laws of motion with quadratic drag in horizontal and vertical motion results in a pair of differential equations that can only be solved numerically (there is no general solution).
 
Welcome to PF!

ahadpoolghaha said:
… our teacher said that we should find 53 degrees for theta …

Hi ahadpoolghaha! Welcome to PF! :smile:

(I've always understood that it's 45 degrees without air resistance.)

What formula for air resistance did your teacher tell you to use? :smile:
 
tiny-tim said:
Hi ahadpoolghaha! Welcome to PF! :smile:

(I've always understood that it's 45 degrees without air resistance.)

What formula for air resistance did your teacher tell you to use? :smile:

I'm pretty sure that theta is always lower than 45 degrees with any plausible
formula for air resistance (linear, quadratic)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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