Question

A body of mass $2 \, \text{kg}$ is thrown up vertically with kinetic energy of $490 \, \text{J}$ . If $g=9.8 \, m \, s^{- 2}$ , the height at which the kinetic energy of the body becomes half of the original value, is

• A. $50 \, m$
• B. $25 \, m$
• C. $12.5 \, m$
• D. $19.6 \, m$

Answer 1

Answer: (C)

Initial kinetic energy is $490 \, \text{J}$
Let h be the height at which kinetic energy becomes half of the initial value
$\frac{K E}{2} = \frac{490}{2} = 245 \, \text{J}$
Gain in Potential energy at height h is mgh
This is equal change in kinetic energy
$\Delta K E = \, 490 \, \text{J} - 245 \, \text{J} = 245 \, \text{J} \, $
$m g h = 245 \, \text{J}$
$h = \frac{245}{2 \times 9.8} = 12.5 \, \text{m}$