Question

A person on a roof throws one ball downward and an identical ball upward at the same speed. The ball thrown downward hits the ground with 140 J of kinetic energy. Ignoring air friction, with how much kinetic energy does the second ball hit the ground?

Answer 1

140 J

Step-by-step explanation:

If we Ignore air friction, according to the law of conservation of energy, the mechanical energy, that is, the sum of the potential energy and kinetic energy, it's a constant value:

`E=U+K\\E=mgh+(mv^2)/(2)`

Therefore, for both balls (which have equals mass), the mechanical energy is the same on the roof, since they are thrown with the same speed (v) from the same height (h). When both balls hit the ground, their potential energy is zero (h=0). The mechanical energy is conserved, so both balls have the same kinetic energy in this point.