# Would different intensities affect the droptime of parachutes?

 P: 15 Would the drop time of two parachute be the same if the wind intensity they experience is different? (Assuming everything is the same). I know the fact that the wind caused them to fall sideways would not cause a difference since the vertical height is the same. Could anyone please tell me the answer? Your help would be greatly appreciated. :D
P: 4,043
 Quote by GuMing Would the drop time of two parachute be the same if the wind intensity they experience is different? (Assuming everything is the same).
What do you mean by "wind intensity they experience". The wind relative to them or the true wind over ground?
 Quote by GuMing I know the fact that the wind caused them to fall sideways would not cause a difference since the vertical height is the same.
That is correct for constant uniform wind over ground.
 P: 15 What I mean is the wind that is not from their top and bottom, but from their left, right, front or back.
P: 4,043
Would different intensities affect the droptime of parachutes?

 Quote by GuMing What I mean is the wind that is not from their top and bottom, but from their left, right, front or back.
I understand that the wind is horizontal. My question was if it is air movement relative to the ground, or relative to the parachuted object that is supposed to be different.

Different constant uniform winds relative to the ground will not affect the fall time.
 Homework Sci Advisor HW Helper Thanks P: 9,839 Dropped from a stationary point, there will be a slight slowing early in the descent, before the horizontal movement of the parachutist has matched the windspeed. The reason is the same one that makes a cyclist struggle in a perfect crosswind - that drag is more than proportional to relative speed. If the cyclist has roadspeed V and the crosswind is W, the nett wind is (V^2+W^2)^(1/2). If the drag goes as the cube of speed then the drag force is (V^2+W^2)^(3/2). The component of this that retards the cyclist is ((V^2+W^2)^(3/2))(V/(V^2+W^2)^(1/2)) = (V^2+W^2).V, rather more than the V^3 of calm conditions.

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