A cricket ball of mass and diameter is dropped from rest from a tower of height

PHYSICS · Subjective · Fermi University Physics · Computational Physics · Mechanics & Dynamical Systems · Kinematics and Newtonian dynamics of particles

Computational Physics, Ch A falling body that knows about air: the first simulation

Problem

A cricket ball of mass

m = 0.160\,\mathrm{kg}

and diameter

7.2\,\mathrm{cm}

is dropped from rest from a tower of height

y_0 = 300\,\mathrm{m}

. Air resistance is quadratic,

m\dot v = -mg + bv^2

with downward taken as negative, and

b = \tfrac12\rho_{\rm air}C_{\rm d}A

with

\rho_{\rm air} = 1.2\,\mathrm{kg\,m^{-3}}

C_{\rm d} = 0.5

and

A

the ball's cross-sectional area found from its diameter. Take

g = 9.81\,\mathrm{m\,s^{-2}}

and ignore spin and any change of shape. A simulation of the fall uses the chapter's update rule, carrying the height forward with the velocity the step started with, at a step

\Delta t = 1.0\,\mathrm{s}

. It reports a height that is still positive at

t=11.0\,\mathrm{s}

y=4.38\,\mathrm{m}

, and that has become

y=-31.44\,\mathrm{m}

t=12.0\,\mathrm{s}

, with velocity

v=-35.84\,\mathrm{m\,s^{-1}}

at that last time. (a) Write the acceleration function

a(y,v)

and the two update equations implied by the equation of motion. (b) Find the terminal speed

v_\infty

predicted by setting

\dot v = 0

. (c) State the bracket that the stopping condition alone gives for the landing (ground-crossing) time, quoting the step

\Delta t

used. (d) Explain why the reported velocity at

t=12.0\,\mathrm{s}

justifies treating the ball as falling at essentially constant speed during the last step, and use that to give a single interpolated crossing time for this

\Delta t = 1.0\,\mathrm{s}

run. Say in one sentence why that number is not yet the landing time of the model, and what rerun would test it.

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