Question

A body of mass 2 kg is thrown up vertically with K.E. of 490 joules. If the acceleration due to gravity is 9.8$m / s ^ { 2 }$, then the height at which the K.E. of the body becomes half its original value is given by

• A. 50 m
• B. 12.5 m
• C. 25 m
• D. 10 m

Answer 1

Answer: (B)

12.5 m

Answer 2

Let h is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy

∴ mgh = $\frac { 490 } { 2 }$⇒ $2 \times 9.8 \times h = \frac { 490 } { 2 }$ ⇒