An ambulance needs to be delivered to a remote town devastated by a major earthquake. All roads leading into the town are blocked due to the earthquake and the ambulance can only be rushed to the area by airlift. The ambulance will be pushed out of a military cargo jet at 3000 m altitude
and rescue staff need to find out what kind of parachute is needed for this mission. The drag force is given by the approximate formula: F = ¼ ρAv2, where ρ is the density of air and ρ = 1.2 kg/m3, A is the area of the cross-section of the parachute perpendicular to the motion and v is the
What should the diameter of the parachute be so that the ambulance can land safely?
Notice that you have to make assumptions on what is approximately a safe landing velocity and what is a weight of a typical ambulance.
Make sure you justify all your assumptions.
2. Relevant equations
mg = (1/4)pAv^2
3. The attempt at a solution
I isolated A (area):
A = (4mg)/(pv^2)
I replaced A with the formula for the area of a circle:
(1/4)∏d^2 = (4mg)/(pv^2)
I isolated d (diameter):
d = √(16mg)/(1.2∏v^2)
Assuming the ambulance weighs a relatively low 3500kg, and that the ambulance lands at 6 m/s (airbags are typically deployed at around a collision speed of 6.2 m/s), I get a parachute diameter of approximately 63.6 m
The problem is my teacher says that value is far, far too big. What am I missing?
Thank you for your time.