You release a 0.57 kg basketball from a height of 10 m. Neglecting air friction, when the ball is 5 m from the ground, its ______ will be ______. A. kinetic energy; 0 J B. potential energy; 5 J C. kinetic energy; 27.9 J D. potential energy; 55.9 J
The original potential energy is given by [tex]E_{potential}=mgh=0.57 \times9.8 \times10=55.9J[/tex] When it's 5m from the ground, part of the original potential energy is now kinetic energy. The potential energy is now [tex]E_{potential}=mgh=0.57 \times9.8 \times5=27.9J[/tex] and the kinetic energy is the original energy minus the current potential energy 55.9J-27.9J=27.9J Answer C is the correct one.
