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Estimate the speed of the parachutist descending at the constant velocity with open parachute. Assume the area of the parachute, A = 30 m^2, mass of a person, m = 100 kg, and density of air, roh = 1 kg/m^3. Enter the answer limited to the second decimal place.
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 -

    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

    For more about the formula, see

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