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If I've a free falling object from a known height above sea level. The object has longitudinal streamlined body so aerodynamically efficient, under 0.09 drag coefficient as heading down with its nose.
How can I calculate the wind effect of 10 m/s on such object supposing that wind is hitting it from side - wind is from East from example - perpendicular to its falling axis? This axis has sectional area of 145 $cm^2$ and the object weights 1500 grams.

I'm newbie on physics, I did my best researching but found no simple solution equation for that.

Dawood
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  • Seems more Physics.SE related. However read this before posting. – user202729 Oct 14 '17 at 08:04
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    Perpendicular effects only affect horizontal motion so the will modify the equation for the horizontal motion with a constant velocity offset – Triatticus Oct 14 '17 at 09:06
  • @user202729 It's not a homework questions I'm honest about it. – Dawood Oct 14 '17 at 12:01
  • @Triatticus Can you give some hints or online resources about this topic? – Dawood Oct 14 '17 at 12:02
  • try searching "projectile motion with wind" – Nick Pavlov Oct 14 '17 at 13:05
  • If the object has a non-negligible cross section for the perpendicular wind then the simplified equation for the drag force is $\frac{1}{2}C\rho A v^2$ where $C$ is the drag coefficient, $\rho$ the density of the moving air, $A$ the cross sectional area of the object, and $v$ the objects velocity with respect to the air (here the horizontal velocity as that is the effect you're asking about). Then you just have to figure out its acceleration due to the force of the wind on the object – Triatticus Oct 14 '17 at 17:53
  • @Triatticus Thanks a lot, so $v$ is not the velocity of the wind in this case, 10 mps ? I've reached this equation and the acceleration after searching as recommended by NickPavlov But the confusing part is $v$ . I found my the equation would be: $Dt = \frac{1}{2} Fd / m$ as the initial velocity is zero in my case. $Dt$ is the distance pushed in specific time. $m$ is the mass. $Fd$ is the drag force as shown by your equation $\frac{1}{2}C\rho A v^2$ . Is this correct? If not correct me please. And also as mentioned explain what $v$ is exactly in my case? – Dawood Oct 15 '17 at 10:32

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