question_answer

A body of mass 2 kg is thrown up vertically with a kinetic energy of 490 J. If the acceleration due to gravity is \[\text{9}.\text{8 m}/{{\text{s}}^{\text{2}}}\], the height at which the KE of the body becomes half of the original value is ?????

Answer:

P E at a height \[h=\frac{1}{2}KE\] \[\Rightarrow mgh=\frac{490}{2}\] \[\Rightarrow 2\times 9.8\times h=\frac{490}{2}\] \[\Rightarrow h=\frac{490}{2\times 2\times 9.8}=12.5m\]