A body is moving at 12 m/s. What would be the vertical height that this body could climb to before it comes to rest?

To determine the vertical height that a body moving at an initial velocity can climb before coming to rest, we can use the principle of energy conservation, specifically the relationship between kinetic energy and gravitational potential energy.

The kinetic energy (KE) of the body at the beginning can be expressed as:

[ \text{KE} = \frac{1}{2} mv^2 ]

where ( m ) is the mass of the body and ( v ) is its initial velocity (in this case, 12 m/s). As the body ascends, this kinetic energy is converted into gravitational potential energy (PE), given by:

[ \text{PE} = mgh ]

where ( g ) is the acceleration due to gravity (approximately 9.81 m/s²) and ( h ) is the height.

At the point where the body comes to rest at the maximum height, the kinetic energy will have been fully converted into potential energy:

[ \frac{1}{2} mv^2 = mgh ]

The mass ( m ) can be cancelled out on both sides, leading to:

[ \frac{1}{2} v^2 = gh ]

Rearranging this formula to solve for