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misUnderstand Gragh of ball moving with resistence of the ai

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ArezList

Posted: Wed Jul 09, 2008 8:31 pm Post subject: misUnderstand Gragh of ball moving with resistence of the ai

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Joined: 22 Mar 2008
Posts: 58

Hi all

Here is a question:

a ball with the mass M is thrown with the angle of Δ,and a the initial vleocity of V, the air resistenec is propositonal with V(2){V square}, namely: Resistence= kV(2), K is a constant. The acceleration of gravitivity is g (m/s2)

Question: what is the Velocity and displacement chart , compare to the graph without air resistence?

Undersatnd the question ? sorry for my english.... By the way can someone analyse it with equations shown, thank you very much!

Harold14370

Posted: Thu Jul 10, 2008 2:01 am Post subject:

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Try this article.

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

Cold Fusion

Posted: Thu Jul 10, 2008 2:41 am Post subject:

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Joined: 10 Apr 2007
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While the ball would fall a little faster (depending on various factors in an atmosphere....whether it is dimpled, its weight, shape) in a vacuum, it would also have less resistance to travel in the horizontal direction. So, it could potentially land in a very similar spot to the one in an atmosphere.

Quote:

Undersatnd the question ? sorry for my english....

And yet you still give him a wikipedia article? Surprised

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Last edited by Cold Fusion on Thu Jul 10, 2008 4:15 pm; edited 1 time in total

ArezList

Posted: Thu Jul 10, 2008 4:29 am Post subject:

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Joined: 22 Mar 2008
Posts: 58

ok that's fine but

which will take longer time rising(T1) or falling(T2),

then if we define that without air resistence the time of ball rising is T,
what's the relations between T1 and T, and also T2 and T?

By the way does my english too poor, or already poor enough?

Harold14370

Posted: Thu Jul 10, 2008 5:21 am Post subject:

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ArezList wrote:

Did you look at the chart in the wikipedia article? It has the answer to your question.

JaneBennet

Posted: Thu Jul 10, 2008 8:30 pm Post subject:

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Joined: 06 Apr 2008
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Harold, the Wikipedia article is assuming that air resistance is proportional to the velocity; this problem in this thread is assuming is assuming itâ€™s proportional to the square of the velocity. The two problems are different.
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Last edited by JaneBennet on Fri Jul 11, 2008 5:47 am; edited 2 times in total

JaneBennet

Posted: Thu Jul 10, 2008 10:05 pm Post subject: Re: misUnderstand Gragh of ball moving with resistence of th

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Joined: 06 Apr 2008
Posts: 801

ArezList wrote:

a ball with the mass M is thrown with the angle of Δ,and a the initial vleocity of V, the air resistenec is propositonal with V(2){V square}, namely: Resistence= kV(2), K is a constant. The acceleration of gravitivity is g (m/s2)

Question: what is the Velocity and displacement chart , compare to the graph without air resistence!

From what Iâ€™ve worked through of the question so far, it looks like the answers are not going to be pretty-looking â€¦ Shocked

You have

, where

and

are respectively the horizontal and vertical components of the velocity.

Then $\dfrac{dv_x}{dt}$ and $\dfrac{dv_y}{dt}$ are respectively the horizontal and vertical components of the acceleration. The forceNow consider the horizontal and vertical forces on the ball:

Horizontally: $M\dfrac{dv_x}{dt}=-kv_x^2$

Vertically: $M\dfrac{dv_y}{dt}=\left\{\begin{array}{ll}-Mg-kv_y^2 & \mbox{going up}\\\\-Mg+kv_y^2 & \mbox{coming down}\end{array}$

Each of the differential equations to be solved is of the variables-separable kind.

For the horizontal case, the solution is $v_x=\dfrac{M}{C_x+kt}$ , where C_x

is a constant. At time t=0

, $v_x=V_0\cos{\Delta}$ (assuming the angle is with respect to the horizontal); therefore $C_x=\dfrac{M}{V_0\cos{\Delta}}$ . Hence $v_x=\dfrac{MV_0\cos{\Delta}}{M+(kV_0\cos{\Delta})t}$ .

Similarly, solve the differential equations for v_y

, then substitute v_x

and

into $v=\pm\sqrt{v_x^2+v_y^2}$ to get your velocityâ€“time relations.

For the displacementâ€“time relations, use $v=\dfrac{ds}{dt}$ .
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