Asked by Anon
A particle of mass 100 gm is thrown vertically upwards with a velocity of 7 m/s. Air resistance is also acting on the particle according to the equation F = -0.2v^2. ( F = Force, v = velocity of the particle ). Find how much high the ball rises.
Please answer it.
Thanks in advance.
Please answer it.
Thanks in advance.
Answers
Answered by
Steve
I'm having a bit of trouble with F. Force is in units of N = kg-m/s^2
Having v^2 there makes the units m^2/s^2. How do you get rid of the extra meters?
Anyway, when you fix that, just remember that F = ma, so a = F/m. Just add that to the -9.8 m/s^2 of gravity, and find the vertex of the resulting parabola.
Having v^2 there makes the units m^2/s^2. How do you get rid of the extra meters?
Anyway, when you fix that, just remember that F = ma, so a = F/m. Just add that to the -9.8 m/s^2 of gravity, and find the vertex of the resulting parabola.
Answered by
bobpursley
This is a bit tricky.
air resistance is indeed dependent on velocity squared.
initial KE=final height- friction work
1/2 m 7^2=mgh - friction
friction losses:
during falling, F=ma=m(dv/dt)
but we know F=mg-kv^2 or
m(dv/dt)=mg-kv^2
so now you have a differential equation in v.
v'=g-k/m v^2
This is not a trivial problems. See page 10 at https://prettygoodphysics.wikispaces.com/file/view/DifferentialEquations.pdf
air resistance is indeed dependent on velocity squared.
initial KE=final height- friction work
1/2 m 7^2=mgh - friction
friction losses:
during falling, F=ma=m(dv/dt)
but we know F=mg-kv^2 or
m(dv/dt)=mg-kv^2
so now you have a differential equation in v.
v'=g-k/m v^2
This is not a trivial problems. See page 10 at https://prettygoodphysics.wikispaces.com/file/view/DifferentialEquations.pdf
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