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 - 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 7:58pm
I answered it. Why are you posting it again?