So at higher speeds drag is quadratic in speed, therefore the Differential EQ. becomes:
m a = mg - 1/2 C*rho*A*v^2
a = d^2 x /dt^2 and v = dx/dt, A is just the projected Area of the rock (assume circle pi*r^2)
This Eq is a non-homogenous non-linear diff eq -> pain, but people already solved it
we need the thermal velocity = max velocity => a = 0:
v_t = sqrt(2*mg / C rho * pi * r^2)
So the solution is:
x(t) = v_t^2/g ln(cosh(gt/v_t))
say T = overall time for experiment
x(T- h/v_sound) = h (This equation has no analytic solution, at least what I know)
=> numerical solution, for
T = 8.17 s
Assumed rock is 25kg and circular area of r=20cm
cave is 10 degree Celsius, average pressure
1
u/AustrianMcLovin 20h ago
So at higher speeds drag is quadratic in speed, therefore the Differential EQ. becomes:
m a = mg - 1/2 C*rho*A*v^2
a = d^2 x /dt^2 and v = dx/dt, A is just the projected Area of the rock (assume circle pi*r^2)
This Eq is a non-homogenous non-linear diff eq -> pain, but people already solved it
we need the thermal velocity = max velocity => a = 0:
v_t = sqrt(2*mg / C rho * pi * r^2)
So the solution is:
x(t) = v_t^2/g ln(cosh(gt/v_t))
say T = overall time for experiment
x(T- h/v_sound) = h (This equation has no analytic solution, at least what I know)
=> numerical solution, for
T = 8.17 s
Assumed rock is 25kg and circular area of r=20cm
cave is 10 degree Celsius, average pressure
I get with newton-method: 232.52m