91Ó°ÊÓ

• $\mathrm{Initial}\mathrm{Velocity}\left({\mathrm{v}}_{\mathrm{i}}\right)=0\mathrm{m}/\mathrm{s}$.
• $\mathrm{Displacement}\left(\mathrm{y}\right)=1700\mathrm{m}.$
• $\mathrm{Acceleration}\left(\mathrm{a}\right)=\mathrm{Gravitational}\mathrm{Acceleration}=9.8\mathrm{m}/{\mathrm{s}}^2.$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. As the air resistance was neglected in this problem, the acceleration will be constant in the given situation. The raindrops would be acted upon by only gravitational acceleration.

Using the kinematic equation, the velocity of the drop can be found.

The final velocity in the kinematic equation is given by,

${{\mathrm{v}}_{\mathrm{f}}}^2={{\mathrm{v}}_0}^2+2\mathrm{ay}\left(\mathrm{i}\right)$

Here ${\mathrm{v}}_0$ is the initial velocity, a is acceleration as y is the displacement.

As there is no air resistance present, acceleration will be constant throughout the problem.Assuming the downward direction is positive and the upward is negative.

Using notation given in equation (i) the velocity can be written as

Substitute the given values in the above equation.

$$\begin{gathered} {\mathrm{v}}^2=0+\left(2×9.8\mathrm{m}/{\mathrm{s}}^2×1700\mathrm{m}\right) \\ {\mathrm{v}}^2=33320{\mathrm{m}}^2/{\mathrm{s}}^2 \\ \mathrm{v}=182.53 \\ ≈183\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of the drop is$183\mathrm{m}/\mathrm{s}$ .

From part (a) we can see that the speed of raindrops is quite high. The mass of the raindrop is very small. But in a rainstorm, there will be several raindrops hitting the person at the same time. Even a small mass hitting a person with such a high velocity will be extremely dangerous. Therefore, it is unsafe to walk outside during a rainstorm.