Question

Exercise L.0 Constants A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 225 m/ and feels no appreciable air resistance. When it has reached a height of 600 m its engines suddenly fail so that the only force acting on it is now gravity You may want to review (Pages 50-53) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Up-and- down motion in free fall Her 0

Answer 1

V^2 - u^2 = 2ah 0 - 51.96^2 = -2 times 9.8 times s s = 138m so max imum height reached = 138m b) elapsed time s = ut + 1/2at^2 - 600 = 51.96 times t + 0.5 times (-9.8)t^2 4.9t^2 - 51.96t - 600 = 0 t = 17.6sec c)velocity before crash is v^2 = u^2 + 2as v^2 = 51.96^2 + 2 times 9.8 times 600 v = 120mls d) total time the rocket in air v = u + at^1 51.96 = 0 + 2.25 times t^1 t^1 = 23.1sec t_tot = t + t^1 = 17.6 + 23.1 t_tot = 40.7sec \r\n initial velocity (u) = 0 t = 1.90s a = g = 9.8mls^2 he final velocity of bricks when reaches ground v = u + at v = 0 + 9.8 times 1.90 v = 18.62mls distan ce covered by the bricks s = ut + 1/2at^2 s = 0 + 0.5 times 9.8 times 1.90^2 s = 17.7m