Question

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MATLAB question

Answer 1

For first 0.15 sec because F_net = 16 - mg = ma [g = 10 m/sec^2] implies a = 16 - 0.05 times 10/.05 = 15.5/.05 = 310 m/sec^2 then velocity of particle After 0.15 sec (Let)v_1 = 0 + a t_1 implies v_1 = 46.5 m/sec dist travel by rocket In 0.15 sec d_1 = 1/2 times 310 times (.15)^2 = 3.48 m After t = 0.15 sec/v_2 = 0 because v_2 = v_1 g t_2 v/g = t_2 implies t_2 46.5/10 = 4.65 sec dist travel by rocket In 4.65 sec d_2 = v_1 t - 1/2 g t^2 d_2 = 46.5 times 4.65 - 1/2 times 10 times (4.65)^2 d_2 = 108.1125 m For down ward motion because v_3 = v_2 + g t_3 t_3 = 2 sec dist travel by rocket In 2 sec d_3 = 1/2 g t^2_3 p = 1/2 times 10 times 4 = 20 m because d_1 + d_2 = d_3 + d_4 d_4 = 91.60 m and t_4