A fruit bat falls from the roof of a cave. We know that her potential energy was 600J, and that the roof of the cave was 12 meters high. What is her velocity when she hits the ground?

Answer 1

Let PE =mgh and KE = mv²/2 At top of the cave the bat has potential energy but not kinetic energy. At the bottom of the cave the bat now has kinetic energy but no potential. So PE_i + KE_i= PE_f +KE_f So now using the forms for PE and KE mg12 + 0 = 0 + mv²/2 The m's cancel and v = √(24* 9.8) m/sec and v = 15.34 m/sec.