A boy is sitting on a swing at a maximum height of 5m above the ground. When the swing passes through the mean position which is 2m above the ground its velocity is approximately

The correct option is 7.6 m/s.

To find the velocity of the boy at the mean position, we can use the law of conservation of mechanical energy. Since there is no friction or air resistance mentioned, the total mechanical energy (potential energy + kinetic energy) of the swing-boy system remains constant throughout the motion.

Let:
- $h_{\max }$ be the maximum height of the swing above the ground, which is $5\text{m}$.
- $h_{\mathrm{mean}}$ be the height of the swing at the mean position, which is $2\text{m}$.
- $v$ be the velocity of the swing at the mean position.
- $g$ be the acceleration due to gravity, approximately $9.8{\text{m/s}}^2$.
- $m$ be the mass of the boy.

At the maximum height, the swing momentarily comes to rest, meaning its velocity is zero. Therefore, its energy is entirely gravitational potential energy:

$E_{\mathrm{initial}}=mg{h}_{\max }$

At the mean position, the swing is at a lower height and has gained velocity. Its total mechanical energy is the sum of its potential energy at this height and its kinetic energy:

$E_{\mathrm{final}}=mg{h}_{\mathrm{mean}}+\frac{1}{2}m{v}^2$

According to the law of conservation of energy, the initial energy equals the final energy:

$mg{h}_{\max }=mg{h}_{\mathrm{mean}}+\frac{1}{2}m{v}^2$

We can divide the entire equation by the mass $m$, showing that the velocity is independent of the boy's mass:

$g{h}_{\max }=g{h}_{\mathrm{mean}}+\frac{1}{2}{v}^2$

Rearranging the equation to solve for $v^2$:

$\frac{1}{2}{v}^2=g(h_{\max }-h_{\mathrm{mean}})$

$v^2=2g(h_{\max }-h_{\mathrm{mean}})$

Now, substitute the given values into the formula:

$v^2=2\times 9.8\times (5-2)$

$v^2=2\times 9.8\times 3$

$v^2=58.8$

Taking the square root of both sides to find the velocity $v$:

$v=\sqrt{58.8}\approx 7.67\text{m/s}$

This value is approximately $7.6\text{m/s}$, matching the correct option.