A105-kg basketball player crouches down 0.400 m while waiting to jump. after exerting a force on the floor through this 0.400 m, his feet leave the floor and his center of gravity rises 0.950 m above its normal standing erect position. (a) using energy considerations, calculate his velocity when he leaves the floor. (b) what average force did he exert on the floor? (do not neglect the force to support his weight as well as that to accelerate him.) (c) what was his power output during the acceleration phase?

A) acceleration. this seems to be the one that makes sense.

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.