Terminal velocity!
An object falling from rest in air is subject not only to gravitational force but also to air resistance. Assume air resistance is proportional to the velocity with constant of proportionality as k >0, and acts in a direction opposite to motion. If the velocity cannot exceed A m/s then, what is the value of 1 0 A k ? (Take g = 9 . 8 m/s 2 )
The answer is 98.
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2 solutions
Shouldn't the mass be given or mentioned in the problem? The thought the force from air resistance was ma=kv, not ma=mkv. I just ended up assuming the mass was 1 to solve the problem because without that, it seemed unsolvable.
Assuming the object's mass is 1 kg, we can model its free-fall motion according to the ODE:
m v ′ ( t ) = m g − k v ( t ) ; v ( 0 ) = 0
which separates & integrates according to:
g − ( k / m ) v ( t ) d v ( t ) = d t ;
or ( − k m ) ⋅ ln [ g − ( k / m ) v ( t ) ] = t + C
which the initial condition v ( 0 ) = 0 yields: C = − k m ⋅ ln ( g ) . Solving for v ( t ) ultimately gives us:
v ( t ) = k m g ⋅ ( 1 − e − k t / m ) .
As t → ∞ , then A = lim t → ∞ v ( t ) = k m g , and A k = m g . This finally gives us 1 0 A k = 1 0 m g = 1 0 ( 1 ) ( 9 . 8 ) = 9 8 .
Let the object have mass=m; After attaining the terminal velocity, net force acting on the body is zero. i.e gravitational force(downwards)=air resistance(upwards) i.e mg=mkA which gives us; g=kA therefore 10kA=10g; =98