Your friend has climbed a tree to a height of 6.00 m. You throw a ball vertically up to her and it is traveling at 5.00 m/s when it reaches her. What was the speed of the ball when it left your hand if you released it at a height of 1.10 m?

Answer 1

Assume the total energy (kinetic and gravitational potential) is the same at both heights.

M*g *1.1 + (M/2)V1^2 = M*g*6.00 + (M/2)V2^2

g*(6 - 1.1) = (1/2)(5^2 -V2^2)

Solve for the velocity at 6 meter height, which I have called V2.

Answer 2

is g in this equation 9.8 or -9.8?