Question

A ball is thrown straight upwards at a speed of 32.3 m/s and from a height of 1.5 m above the ground. How fast is it traveling when it hits the ground? (m/s)

Answer 1

Initial Kinetic energy = (1/2) m v^2

= (1/2)m(32.3)^2
Initial potential energy = m g h
= m (9.8)(1.5)
final kinetic energy = (1/2) m v^2
final potential energy = 0
final = initial
(1/2) v^2 = (1/2)(32.3)^2 + 9.8(1.5)