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November 27, 2014

November 27, 2014

Posted by **Anonymous** on Wednesday, February 23, 2011 at 6:51pm.

Hint 1: air pushing on the parachute exerts the resistive force which matches (equal and opposite) the gravitational force on the person. Hint 2: speed of the air striking the parachute is equal to the speed of the parachutist. Hint 3: force exerted by the air on parachute = (impulse exerted by air)/time interval during which the air strikes the parachute Hint 4: mass of air molecules striking the parachute can be found as mass of the air molecules in the cylinder of air below the parachute Hint 5: height of this cylinder = velocity of air multiplied by the time interval during which it is stopped by the parachute

* physics - drwls, Monday, February 21, 2011 at 12:07pm

I find the hints confusing. Here is the formula you need.

M*g = (1/2)*rho*A*Cd*V^2

Solve for the limiting velocity, V

*Cd is the dimensionless "drag coefficient", which is about 1.5 for a hemispherical parachute shape.

*rho is the density of air

*A is the projected area of the parachute (pi R^2), (not the surface area)

*g is the acceleration of gravity

* M is the mass

- physics -
**drwls**, Wednesday, February 23, 2011 at 7:58pmI answered it. Why are you posting it again?

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