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- Thread starter blade_chong
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Yes, air resistance will cut down the speed at which the object falls.

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Yes, air resistance will cut down the speed at which the object falls.

Air resistance will reduce the time needed to reach maximum height and increase the time needed to return from maximum height to original position. I just want to know whether if the reduce the in time needed to reach maximum height will be compensated by the increase the time needed to return from maximum height to original position such that the total time taken equals to T. I am confuse about this part. Pls help me to clarify. Thanks

- #4

uart

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Interestingly (for the same initial velocity) the air resistance projectile always returned slightly sooner than the ideal resistance free projectile. I was able to show that this is the case for all positive values of parameters (mass, gravity and air resistance coefficient). The maths was a bit tedious for me to write up now but I will post it later if you’re interested.

BTW. When the air resistance coefficient is relatively low then the times where very similar, as expected, with the time of the projectile with air resistance included being only very slightly quicker.

- #5

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Interestingly (for the same initial velocity) the air resistance projectile always returned slightly sooner than the ideal resistance free projectile. I was able to show that this is the case for all positive values of parameters (mass, gravity and air resistance coefficient). The maths was a bit tedious for me to write up now but I will post it later if you’re interested.

BTW. When the air resistance coefficient is relatively low then the times where very similar, as expected, with the time of the projectile with air resistance included being only very slightly quicker.

Thanks uart for your explaination about projectile motion. But i wasn't referring to projectile motion. I was referring throwing a ball being thrown up in the air, in which its path up to its maximum height and return journey is always perpendicular to the ground at all time. Can anyone help to explain my doubt? Thanks

- #6

uart

Science Advisor

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Thanks uart for your explaination about projectile motion. But i wasn't referring to projectile motion. I was referring throwing a ball being thrown up in the air, in which its path up to its maximum height and return journey is always perpendicular to the ground at all time. Can anyone help to explain my doubt? Thanks

Yes I understood your question and that's the case I analysed, verticle up and down with no sideway motion. There's no reason why we can not still refer to this object as a projectile ok.

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Yes I understood your question and that's the case I analysed, verticle up and down with no sideway motion. There's no reason why we can not still refer to this object as a projectile ok.

dang..sry abt that...is it possible to show me the equations? equations tell the whole story =D..thanks

- #8

uart

Science Advisor

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Ok here’s the working.

The differential equation of motion with a linear air resistance term included is (using the notation y' for the time derivative and y'' the second derivative) :

[tex]y'' = - ky' - g[/tex]

This DE is pretty easy to solve and gives the following equation for position y (height) versus t (time).

[tex]y = A e^{-kt} + B - (g/k) t[/tex]

Taking y(0)=0 and y'(0)=u gives the following values for the free constants A and B.

[itex]A = -(u/k+g/k^2)[/itex] and [itex]B=-A[/itex]

So finally we have,

[tex]y = (u/k + g/k^2) (1 - e^{-kt}) - (g/k)t[/tex]

Ideally we'd now like to solve this equation and find the value of time "t" when "y" returns to zero. Unfortunately it’s a transcendental equation which means that the solution can't be expressed in terms of elementary functions.

So rather than trying to solve exactly when "y" is zero, a simpler approach is to just substitute [itex]t=2u/g[/tex] into the equation and determine whether the resulting value of "y" is positive or negative. If the resulting "y" is positive then this projectile takes longer to return to ground than does one which doesn't experience air resistance, and visa versa. The reason for this conclusion is that [itex]t=2u/g[/tex] is precisely the time taken for the no air resistance projectile to return to zero, so we can test the relative time to return to ground by simply comparing the position of the air resistance included projectile at this value of time.

Substituting [itex]t=2u/g[/tex] we get :

[tex]y = g/k^2 - u/k - (u/k + g/k^2) e^{-2uk/g}[/tex]

At this point we'd like to show that this value of "y" is either always positive or always negative for any feasible values of the parameters u,g and k. This last part of the problem was a little tedious, there might have been an easier way but below is my proof that "y" is always negative when t = 2u/g.

Firstly note that the equation for "y" is of the form :

[tex]y = a - b - (b + a) e^{-2b/a}[/tex]

Where constants "a" and "b" are defined from the earlier equation in the obvious way.

Manipulating this a bit we get :

[tex]y = -2b + (a + b) (1 - e^{-2b/a})[/tex]

Finally, by dividing both sides by "a" and we can make this an equation in just one variable.

[tex]y/a = - 2c + (1 + c) (1 - e^{-2c})[/tex]

Where [itex] c = \frac{b}{a} = \frac{u k}{g}[/itex]

Denote this function as [itex]f(c)[/itex], that is :

[tex] f(c) = -2c + (1+c)(1-e^{-2c}) [/tex]

It’s easy to show the following about this function.

1. f(0) = 0

2. f(c) tends to negative infinity as c goes to infinity.

3. The derivative f'(c) is zero at c=0 but negative for all c>0.

From the above we can deduce that f(c) is negative for all c>0, and therefore that the value of "y" is negative at t=2u/g for all positive values of the parameters. Essentually this proves that the air resistance included projectile always returns to ground sooner than one in which air resistance is not included.

The differential equation of motion with a linear air resistance term included is (using the notation y' for the time derivative and y'' the second derivative) :

[tex]y'' = - ky' - g[/tex]

This DE is pretty easy to solve and gives the following equation for position y (height) versus t (time).

[tex]y = A e^{-kt} + B - (g/k) t[/tex]

Taking y(0)=0 and y'(0)=u gives the following values for the free constants A and B.

[itex]A = -(u/k+g/k^2)[/itex] and [itex]B=-A[/itex]

So finally we have,

[tex]y = (u/k + g/k^2) (1 - e^{-kt}) - (g/k)t[/tex]

Ideally we'd now like to solve this equation and find the value of time "t" when "y" returns to zero. Unfortunately it’s a transcendental equation which means that the solution can't be expressed in terms of elementary functions.

So rather than trying to solve exactly when "y" is zero, a simpler approach is to just substitute [itex]t=2u/g[/tex] into the equation and determine whether the resulting value of "y" is positive or negative. If the resulting "y" is positive then this projectile takes longer to return to ground than does one which doesn't experience air resistance, and visa versa. The reason for this conclusion is that [itex]t=2u/g[/tex] is precisely the time taken for the no air resistance projectile to return to zero, so we can test the relative time to return to ground by simply comparing the position of the air resistance included projectile at this value of time.

Substituting [itex]t=2u/g[/tex] we get :

[tex]y = g/k^2 - u/k - (u/k + g/k^2) e^{-2uk/g}[/tex]

At this point we'd like to show that this value of "y" is either always positive or always negative for any feasible values of the parameters u,g and k. This last part of the problem was a little tedious, there might have been an easier way but below is my proof that "y" is always negative when t = 2u/g.

Firstly note that the equation for "y" is of the form :

[tex]y = a - b - (b + a) e^{-2b/a}[/tex]

Where constants "a" and "b" are defined from the earlier equation in the obvious way.

Manipulating this a bit we get :

[tex]y = -2b + (a + b) (1 - e^{-2b/a})[/tex]

Finally, by dividing both sides by "a" and we can make this an equation in just one variable.

[tex]y/a = - 2c + (1 + c) (1 - e^{-2c})[/tex]

Where [itex] c = \frac{b}{a} = \frac{u k}{g}[/itex]

Denote this function as [itex]f(c)[/itex], that is :

[tex] f(c) = -2c + (1+c)(1-e^{-2c}) [/tex]

It’s easy to show the following about this function.

1. f(0) = 0

2. f(c) tends to negative infinity as c goes to infinity.

3. The derivative f'(c) is zero at c=0 but negative for all c>0.

From the above we can deduce that f(c) is negative for all c>0, and therefore that the value of "y" is negative at t=2u/g for all positive values of the parameters. Essentually this proves that the air resistance included projectile always returns to ground sooner than one in which air resistance is not included.

Last edited:

- #9

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- 0

Ok here’s the working.

The differential equation of motion with a linear air resistance term included is (using the notation y' for the time derivative and y'' the second derivative) :

[tex]y'' = - ky' - g[/tex]

This DE is pretty easy to solve and gives the following equation for position y (height) versus t (time).

[tex]y = A e^{-kt} + B - (g/k) t[/tex]

Taking y(0)=0 and y'(0)=u gives the following values for the free constants A and B.

[itex]A = -(u/k+g/k^2)[/itex] and [itex]B=-A[/itex]

So finally we have,

[tex]y = (u/k + g/k^2) (1 - e^{-kt}) - (g/k)t[/tex]

Ideally we'd now like to solve this equation and find the value of time "t" when "y" returns to zero. Unfortunately it’s a transcendental equation which means that the solution can't be expressed in terms of elementary functions.

So rather than trying to solve exactly when "y" is zero, a simpler approach is to just substitute [itex]t=2u/g[/tex] into the equation and determine whether the resulting value of "y" is positive or negative. If the resulting "y" is positive then this projectile takes longer to return to ground than does one which doesn't experience air resistance, and visa versa. The reason for this conclusion is that [itex]t=2u/g[/tex] is precisely the time taken for the no air resistance projectile to return to zero, so we can test the relative time to return to ground by simply comparing the position of the air resistance included projectile at this value of time.

Substituting [itex]t=2u/g[/tex] we get :

[tex]y = g/k^2 - u/k - (u/k + g/k^2) e^{-2uk/g}[/tex]

At this point we'd like to show that this value of "y" is either always positive or always negative for any feasible values of the parameters u,g and k. This last part of the problem was a little tedious, there might have been an easier way but below is my proof that "y" is always negative when t = 2u/g.

Firstly note that the equation for "y" is of the form :

[tex]y = a - b - (b + a) e^{-2b/a}[/tex]

Where constants "a" and "b" are defined from the earlier equation in the obvious way.

Manipulating this a bit we get :

[tex]y = -2b + (a + b) (1 - e^{-2b/a})[/tex]

Finally, by dividing both sides by "a" and we can make this an equation in just one variable.

[tex]y/a = - 2c + (1 + c) (1 - e^{-2c})[/tex]

Where [itex] c = \frac{b}{a} = \frac{u k}{g}[/itex]

Denote this function as [itex]f(c)[/itex], that is :

[tex] f(c) = -2c + (1+c)(1-e^{-2c}) [/tex]

It’s easy to show the following about this function.

1. f(0) = 0

2. f(c) tends to negative infinity as c goes to infinity.

3. The derivative f'(c) is zero at c=0 but negative for all c>0.

From the above we can deduce that f(c) is negative for all c>0, and therefore that the value of "y" is negative at t=2u/g for all positive values of the parameters. Essentually this proves that the air resistance included projectile always returns to ground sooner than one in which air resistance is not included.

thx uart.... =)

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