91Ó°ÊÓ

The given data can be listed below as,

• The roof of the building is$46.0\mathrm{m}$ above the ground.
• The height of the physics professor is $1.8\mathrm{m}$.
• The professor walks at a constant speed of $1.20\mathrm{m}/\mathrm{s}$.

This law states that a body will continue to move in a uniform motion unless an external object acts on the body.

The equation of the displacement gives the distance the professor should be while releasing the egg.

From the Newton’s law of motion, the displacement of the professor can be expressed as:

$\mathrm{y}−{\mathrm{y}}_0=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$

Here, y is the height of the roof that is $46\mathrm{m}$, ${\mathrm{y}}_0$is the height of the professor that is $1.8\mathrm{m}$, u is the initial velocity that is zero as the professor was at rest and t is the time taken by the professor that is also expressed as $\frac{{\text{x}}_{\text{pro}}}{{\text{v}}_{\text{pro}}}$where $x_{\text{pro}}$is the required distance and ${\mathrm{v}}_{\text{pro}}$is the speed of the professor that is $1.2\mathrm{m}/\mathrm{s}$.

Substituting the values in the above equation, we get-

role="math" localid="1655710930907" $\begin{array}{l}46{\text{″¾-1.8″¾=0+4.9″¾/²õ}}^{\text{2}}×\frac{x_{\text{pro}}^2}{v_{\text{pro}}^2}\\ x_{\text{pro}}^2=\frac{44.2\text{″¾}×{\left(\text{1.2″¾/s}\right)}^2}{{\text{4.9″¾/s}}^{\text{2}}}\\ x_{\text{pro}}^{}=3.6\text{″¾}\end{array}$

Thus, the professor should be$3.6\mathrm{m}$ away when I will release the egg.