Energy Transformation Efficiency Problem

[hamza2095]

Homework Statement

A ball that weights 1kg is thrown in the air, the hand being 1m above the ground. The velocity the ball was thrown at was 14m/s and it's maximum height was 6.5m above the ground. What is the efficiency for this energy transformation?

Homework Equations

Do I just get the mechanical energy of when the ball is in the hand (input) and the mechanical energy of when the ball reaches its maximum height (output) and use the standard efficiency equation (My attempt at the solution)? Or do I just act as if the hand is reference point and find the kinetic energy, then the gravitational potential energy (Using 5.5m because hand is 1m above ground) and find the efficiency that way

The Attempt at a Solution

EmBottom = m*g*h + 1/2 (mv^2)
= 1kg*9.8N/kg*1m + 1/2 (1kg(14m/s)^2)
= 107.8 J
EmTop = m*g*h (Just gravitational potential because the velocity at the top is 0, making the kinetic energy 0)
= 1kg*9.8*6.5m
= 63.7J

Efficiency = (output/input) * 100
= (63.7/107.8) * 100
= 59%

 

[haruspex]

The 1m initial height is certainly arbitrary, and is not involved in any transformation. What is transformed is KE into GPE, so only count that.

 

[hamza2095]

The 1m initial height is certainly arbitrary, and is not involved in any transformation. What is transformed is KE into GPE, so only count that.

Damn, was a test question, hopefully i got partial marks lol. Thanks for the help.