Question

The height of a helicopter above the ground is given by h 5 3.00t3, where h is in meters and t is in seconds. At t 5 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground

Answer 1

Answer:

$t \approx 2.212\,s$

Explanation:

The height of the helicopter is given by this function:

$h(t) = 3\cdot t^{3}$

The height at $t = 2\,s$ is:

$h(2\,s)= 3\cdot (2\,s)^{3}$

$h (2\,s) = 24\,m$

The mailbag experiments a free fall, whose equation is:

$y = 24\,m -\frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}$

Time required before the mailbag reaches the ground is:

$24\,m-\frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}=0$

The roots of the second-order polynomial are:

$t_{1} \approx 2.212\,s$

$t_{2} \approx - 2.212\,s$

Only the first root offers an physically reasonable solution. Then, the needed time is:

$t \approx 2.212\,s$