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I am trying to figure out how to put this in and i keep running into issues can anyone help?

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Daniel Bucknavich
Daniel Bucknavich 2022-3-25
here is my code
eq1= sym('(v*(8)==(sqrt(((95*9.81)/0.24)))*(tan(h))*(sqrt(((0.24*9.81)/95)*8)))');
eq2= sym('h*(8)==(95/0.24)*log(cos(h)*(sqrt((0.24*9.81)/95)*8))');
sol = solve(eq1,'v',eq2,'h');

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Sam Chak
Sam Chak 2022-3-25
编辑：Sam Chak 2022-3-25

2 个投票

Hi @Daniel Bucknavich
This is not a direct solution, but it should help you to visualize the physics of skydiving and terminal velocity, and hopefully giving you the idea to finding and when sec.
m = 95;
rho = 0.24;
g = 9.8;
t = 0:0.1:20;
v = sqrt(m*g/rho)*tanh(sqrt(rho*g/m)*t);
y = -(m/rho)*log(cosh(sqrt(rho*g/m)*t)); % purposely added a 'minus' sign
plot(t, v, t, y)

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Torsten
Torsten 2022-3-25

1 个投票

v = sqrt(95*9.8/0.24)*tanh(sqrt(0.24*9.8/95)*8)
y = 95/0.24*log(cosh(sqrt(0.24*9.8/95)*8))

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