A body of mass 2 kg is thrown upward with an energy 490 J. The height at which its kinetic energy would become half of its initial kinetic energy will be [ g = 9.8 m / s² ]

The correct answer is 12.5m.

Step-by-Step Explanation:

1. Identify the given parameters:
Mass of the body, m = 2 kg
Initial energy (which is entirely kinetic energy at the point of projection), K_i = 490 J
Acceleration due to gravity, g = 9.8 m/s²

2. Understand the Conservation of Energy:
According to the principle of conservation of mechanical energy, the total energy (kinetic energy K + potential energy U) remains constant at all points during the motion if air resistance is neglected.

Therefore, at any height h:

$K+U=E_{\text{total}}$

Since the initial total energy is equal to the initial kinetic energy K_i:

$K+U=K_i$

3. Set up the condition for the kinetic energy to become half:
We want to find the height h where the kinetic energy K becomes half of its initial value K_i:

$K=\frac{1}{2}{K}_i$

Substitute this condition into the conservation of energy equation:

$\frac{1}{2}{K}_i+U=K_i$

Solving for the potential energy U at this height:

$U=K_i-\frac{1}{2}$

$U=\frac{1}{2}$

4. Calculate the height h:
The gravitational potential energy at a height h is given by the formula:

$U=m·g·h$

Equating the two expressions for potential energy:

$m·g·h=\frac{1}{2}$

Substitute the given numerical values into the equation:

$2·9.8·h=\frac{490}{2}$

$19.6·h=245$

Solve for h:

$h=\frac{245}{19.6}$

$h=12.5\text{m}$

Thus, the height at which the kinetic energy becomes half of the initial kinetic energy is 12.5m.