Question

I. The mass of baseball is m = 145 g and and diameter D 72.8 mm. For a baseball in air, the drag depends on the square of the speed, f- -c2v|v. The coefficient of friction scales with the diameter D, c2 0.22D2 1.2 x 103 kg/m. For all parts, assume that the ball drops vertically, and plot the acceleration, velocity, and height as a function of time (a) If you were to drop this out of the international space station, that orbits above the Earth at 150 km, what is the maximum velocity if you don't take air resistance into account? (b) Now if you drop this from the ISS but you take c2 into account, what is the maximum velocity? c) Now the coefficient of friction, c2, depends on air density. Recall that air density goes as my-Poe-y/so where yo-10 km. What is the maximum velocity now? Is it at the ground? (d) What if you take the changing gravitational acceleration? Assume that g scales as 9.81 How different is this answer from part (c)?

Answer 1

Given data, m = 145 g D = 72.8 mm. The international space station height = 150 km V_max = squareroot 2 gh (with out air drug) = squareroot 2 times 908 times 150 times 10^3 V_max = 1714.64 m/s Now, consider an air drug, v(t) = v_tan h (gt/vt) m = 145 g = 0.145 kg, c = 1.2 times 10^-3 kg/m Maximum Velocity, V_terminal = squareroot mg/c = squareroot 0.145 times 9.8/1.2 times 10^-3 = squareroot 1.421/0.0012 V_termianl = 34.4 m/sec Depend on air density, if air drug coefficients P = P_0 e^-1/Y_0 The equation of motion is, m du/dt = mg - c prime v^2 c prime = P/p_0 c = ce^-1/y_0 m dv/dt = mg = ce^-Y/Y_0 V^2_max V_max = squareroot mg/c e^1/Y_0 We know, g = 9.81/(1 + 1/Re)^2 The equation of motion, m dV/dt = m times 9.8/(1 + 4/RE)^2 - cv^2 At terminal velocity, dV/dt = 0