## anonymous 2 years ago A stone of mass 0.55 kilograms is released and falls to the ground. Measurements show that the stone has a kinetic energy of 9.8 joules at the point of impact with the ground. What is the stone's velocity when it hits the ground?

1. Shane_B

$KE=\frac{1}{2}mv^2$$9.8J=\frac{1}{2}(0.55kg)v^2$Solve for v.

2. anonymous

thx :)

3. Shane_B

np :)