9. Bonus problem: On Thursday, we have discussed about the special topic about free-falls in re- sistive media. For the more challenging model of dragging force proportional to velocity squared: FD Kv², where x is a constant, solve the problem by completing the following steps. (a) By applying Newton's second law of motion, you get the following equation: dv mg-kv²=m² dt' Explain a bit how this equation is written down. (b) Rewrite the equation given in (a) as the following L-C; dv = g m (2 points) (1) (2 points) and explain a bit how this equation is written down. (c) Equation (1) leads to the solution for v(t), but to solve it, you need a few steps in order to work out the integral at the right hand side of Eq. (1). Consider the following change of variable: K mg Show that the right hand side of Eq. (1) can be written as (3 points) dv Ky m dx = K 8- 1/2 1-x2 (2) m (d) Consider the following integral with c < 1: dx |= Show that the integral can be recast into -8 |= sec ᎾdᎾ Jo by performing another change of variable where 6, fulfills sin 0,= c. (e) Verify by showing x = sin 0, sec Ode In(tan + sec ) = (3) (3 points). (4) d In(tan 0+ sec 0) = sec 0. dᎾ Hint: Use the following basic formulas 2/3 » (0) Inf(0) de f(0) d g(0) (0) (0)-8(0) de f(0) (0) [f(0)]² and basic definitions of tan = sin 0 cose sec 8 cos e All given points from solving bonus problems will be added to the total points of your homework for the entire semester until you reach the maximum, which is planned to be 16 x 100. 2 (f) Using Eq. (4), show that the integral of (3) is given by 1 1+c 1= Hint: From sin =c, you know cos 0 =√1-c²; with sine and cos e, you know tane and sec 0. (g) Show that the above result of (f) can be rewritten as (4 points) I=tanh-c, where tanh is the inverse tanh function. Hint: This means to show tanh(1) = c. (4 points) (h) If you have completed all the above steps, up to here you have shown that dx 1-x =tanh-(c), where c<1. Now, come back to Eq. (2), and use Eq. (5) to show f dv K m dx m -1 tanh Kg Jo mg 8- m (5) (4 points) (i) Finally, substitute the above result into (1), and you should readily derive the desired solution for v(t). END (5 points) 3 3/3