Question

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high above the ground was the package when it was released?

Answer 1

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$y$ - Final height, measured in meters.

$y_{o}$ - Initial height, measured in meters.

$t$ - Time, measured in seconds.

$g$ - Gravitational constant, measured in meters per square second.

(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

$y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}$

Given that $y = 0\,m$, $v_{o} = 15\,\frac{m}{s}$, $g = -9.807\,\frac{m}{s^{2}}$ and $t = 16\,s$, the final height of the package is:

$y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}$

$y_{o} = 1015.296\,m$

The package was released at a height of 1015.296 meters.