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 55.0mph, then the package will break. Neglect drag.At what height would the package hit the ground with the critical speed (in given units)? Could the package be safely dropped from a height of 20m?(Yes or no) Hint: The final speed is given in mph. Convert this to m/s before applying calculations.Heights that are lower than the critical height you calculated are safe, because they allow the package to land with a lower speed.

Answer 1

In order to determine the height at which the package can be dropped, use the following formula:

$v^2=v^2_0+2gy$

where,

v: final speed = 55.0mph

g: gravitational acceleration constant = 9.8 m/s^2

y: height = ?

vo: inital speed of the package = 0 m/s

Convert mph to m/s first:

$55\text{mph}\cdot\frac{1.6\operatorname{km}}{1\text{mile}}\cdot\frac{1000m}{1\operatorname{km}}\cdot\frac{1h}{3600s}=\frac{24.44m}{s}$

Next, solve the equation above for y and replace the values of the other parameters:

$y=\frac{v^2}{2g}=\frac{(\frac{24.44m}{s})^2}{2(\frac{9.8m}{s^2})}=30.48m$

Hence, at 30.48m the package reaches the ground with the critical speed.

Moreover, for 20m the package be safely dropped.