Question

a helicopter is delivering food in an emergency situation, where it is difficult to land. the engineer is tasked with determining what heights the package could be dropped without breaking. she knows that if the package strikes the ground faster than the critical speed of 62.0mph, then the package will break. neglect drag. at what height would the package hit the ground with the critical speed (in given units)? m could the package be safely dropped from a height of 30m? yes

Explanation:

Step1: Convert speed to SI units

First, convert 62.0 mph to m/s. 1 mile = 1609.34 m and 1 hour = 3600 s. So, $v = 62.0\times\frac{1609.34}{3600}\text{ m/s}\approx 27.7\text{ m/s}$.

Step2: Use kinematic - equation

The kinematic equation $v^{2}=v_{0}^{2}+2a\Delta y$ is used. The initial velocity $v_{0} = 0\text{ m/s}$ (package is dropped), acceleration $a = g= 9.8\text{ m/s}^2$, and we want to find $\Delta y$. Rearranging the equation for $\Delta y$ gives $\Delta y=\frac{v^{2}-v_{0}^{2}}{2a}$. Substituting $v_{0} = 0\text{ m/s}$, $v = 27.7\text{ m/s}$, and $a = 9.8\text{ m/s}^2$ into the equation, we get $\Delta y=\frac{27.7^{2}-0^{2}}{2\times9.8}=\frac{767.29}{19.6}\approx39.2\text{ m}$.

Step3: Answer the second - part

We compare the height of 30 m with the calculated critical height. Since $30\text{ m}<39.2\text{ m}$, the package can be safely dropped from 30 m.

Answer:

39.2
yes