Question

30. A mass of 0.40 kg is dropped from rest in a medium of- fering a resistance of 0.2 , where u is measured in meters per second. (a) If the mass is dropped from a height of 25 m, find its velocity when it hits the ground. (b) If the mass is to attain a velocity of no more than 8 m/s, find the maximum height from which it can be dropped. (c) Suppose that the resistive force is kļul, where u is mea- sured in meters per second and k is a constant. If the mass is dropped from a height of 25 m and must hit the ground with a velocity of no more than 8 m/s, determine the coefficient of resistance k that is required. from the

Answer 1

Given m = 0.4 kg and the resistance force F = 0.2 Vertical-bar v Vertical-bar . so the resultant force on the body when it falls down in given by F_R = mg - 0.2 Vertical-bar v Vertical-bar m dv/dt = mg - 0.2 Vertical-bar v Vertical-bar dv/dt + 0.2/m Vertical-bar v Vertical-bar = mg dv/dt + Vertical-bar v Vertical-bar /2 = g dv/dt = Vertical-bar g - Vertical-bar v Vertical-bar /2 Vertical-bar integral^v_0 dv/10 - Vertical-bar v Vertical-bar /2 - inetgral^t_0 dt -1/2 ln Vertical-bar 10 - Vertical-bar v Vertical-bar /2 Vertical-bar = t + c at t = 0 v = 0 -1/2 ln Vertical-bar 10 - Vertical-bar v Vertical-bar /2 Vertical-bar + 1/2 ln Vertical-bar 10 Vertical-bar = t 10 - Vertical-bar v Vertical-bar /2 = 10 (e^-2t) v = 20(1 - e^-2t) dx/dt = 20(1 - e^-2t) integral^x_0 dx = integral^t_0 20 (1 - e^-2t)dt x = + 10 (2 + e^2t) 11 at x = 25 25/10 - 2 = e^-2t 1/2 = e^-2t \r\n Put this value in eq^n 1 V = 20(1 - 1/2) = 10 m/s Given V_max = 8 m/s 8 - 20(1 - e^-2t) 1 - 8/20 = e^-2t = > e^-2t = 3/5 Put this value in equation 11 x =