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Old Sep 24th 2017, 07:24 AM   #1
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Projectile motion

Hello!

I'm a physics student and i have a problem i can't solve. I was wondering if there is anyone here who would be willing to help me solve this?

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. What is the angle of maximum range and how it is dependent on initial velocity if we include air resistance and if the wind is blowing in the horizontal direction of flight?

Thank you so much for your help!
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Old Sep 24th 2017, 07:54 AM   #2
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You should know that the force of gravity is "-mg" so that "-mg= ma" and so a= -g, which is a downward pointing vector. "Air resistance" is typically taken to be proportional to the speed, $\displaystyle \sqrt{v_x^2+ v_y^2}$. If the wind is blowing horizontally with speed w, then we take that into account by replacing $\displaystyle v_y$ with [math]v_y- w[/tex], the horizontal speed relative to the wind.

The components of acceleration are $\displaystyle a_y= -g$, $\displaystyle a_x= \kappa\sqrt{v_x^2+ (v_y- w)^2}$
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Old Sep 25th 2017, 05:27 AM   #3
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Hello.

Yes, but isn't air resistance in the opposite direction of flight? And if we use quadratic air resistance we get two equations:

a_x= -K(v_x^2+(v_y -w)^2 )*cosφ
a_y= -K(v_x^2+(v_y -w)^2 )*sinφ⁡ - g

But i don't know how to solve them.

Last edited by ficku1; Sep 25th 2017 at 05:36 AM.
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Old Sep 25th 2017, 08:39 AM   #4
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Originally Posted by ficku1 View Post
Hello!

I'm a physics student and i have a problem i can't solve. I was wondering if there is anyone here who would be willing to help me solve this?

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. What is the angle of maximum range and how it is dependent on initial velocity if we include air resistance and if the wind is blowing in the horizontal direction of flight?

Thank you so much for your help!
Rules of the forum require you to first make an attempt.

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Old Sep 25th 2017, 10:02 AM   #5
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I am sorry, I only saw the rules after I posted.

I tried to solve the problem but i don't know how to continue, I can't think of any way to solve these two equations if they are correct:

a_x= -K(v_x^2+(v_y -w)^2 )*cosφ
a_y= -K(v_x^2+(v_y -w)^2 )*sinφ⁡ - g
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Old Sep 27th 2017, 07:18 AM   #6
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Sorry i'm a few days late responding to this thread.

1. The wind blow horizontally, so it only has an affect on a_x, not a_y.
2. You can break forces and accelerations into x- and y- components. Hence:

a_x = -K(v_x-w)^2
a_y = -K(v_y)^2 -g for positive values of v_y, and a_y = K(v_y)^2 -g for negative values of v_y.

This assumes that air resistance is proportional to velocity squared, and leads to some pretty complicated mathematics that are best solved using numerical techniques. However, if we can model air resistance as being proportional to velocity, the equations become easier and can be solved with a closed form solution:

a_x = -K v_x + Kw
a_y = -k v_y -g

Or in differential equation form:

$\displaystyle \ddot x + k \dot x + Kw = 0$
$\displaystyle \ddot y + k \dot y +g = 0$

Each of these has a solution of the form $\displaystyle x = Ae^{-ct} + Bt + C$.
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