Estimate the speed of the parachutist descending at the constant velocity with o

Question

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: F_air = F_g = mg.
Hint 2: (initial) speed of the air striking the parachute is equal to the speed of the parachutist, which the problem asks for.
Hint 3: by Newton's 2nd law in Momentum Form, force exerted by the air on parachute F_air = (impulse exerted by air)/time interval during which the air strikes the parachute
Hint 4: mass of air molecules striking the parachute is found as mass of the air molecules in the cylinder of air below the parachute: m_air = roh_air * Volume_cylinder.
Hint 5: height of this cylinder = velocity of air multiplied by the time interval during which it is stopped by the parachute: l_cylinder = v_air*delta_t.

The answer is 5.72 m/s

Explanation / Answer

At terminal velocity, the two forces cancel out So: mg of the parachutist = ma of the air mg = density of air*volume of air *change in velocity of air/sec = ma volume of air is area*height/t = area*velocity height per time is a constant since it is at terminal velocity, so we can use 1 second as the time. This way height per time is the velocity. mg = density*area*velocity*(vo - vf) air is stopped at the end, so (vo - vf) = terminal velocity mg = (1 kg/m^3)*(30 m^2) v^2 v^2 = mg/[(1kg/m^3)30m^2) = 100kg*9.8m/s^2/30m^2 v^2 = 32.6666 m^2/s^2 v = 5.7154 m/s

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