A 0.15-kg ball dropped from a helicopter at a height of 1.0 x 102 m. Show that the speed of the ball is about 44 m/s just before it hits ground (g = 9.8 m/s2 and ignore air

Question

resistance).

Answer 1

gravitational potential energy becomes kinetic energy

1/2 m v^2 = m g h ... v = √(2 g h) = √(2 * 9.8 * 100)