Question

A body of mass 2 kg is thrown up vertically with kinetic energy of 490 j. If g = 9.8 m/s2, the height at which the kinetic energy of the body becomes half of the original value is

Answer 1

energy conservation

K1-K2 =mgh

490-490/2=2×9.8×h

490-245=19.6h

245=19.6h

h =245/19.6

h =12.5 m

Answer 2

K1-K2 = mgh

• 490-490/2 = 2×9.8×h
• 490-245 = 19.6 h
• 245 = 19.6
• h = 245/19.6
• h = 12.5m