A helicopter is rising straight up. the height of a helicopter above the ground is given by h = 3.50t3, where h is in meters and t is in seconds. (so that equation's form tells us the helicopter is clearly not in freefall ... its engines must be working in some way.) at t = 1.75 s, the helicopter releases a small mailbag. how long after its release does the mailbag reach the ground?

First, we need to calculate for the height where the mailbag was dropped.

h = 3.50 * t^3

at t = 1.75 s, h is:

h = 3.50 * (1.75) ^3

h = 18.758 m

Then we can use the equation to solve for time:

h = v0t + 0.5gt^2

where v0 is initial velocity = 0, t = time, g = acceleration due to gravity

18.758 m = 0 + 0.5 * (9.81 m/s^2) * t^2

t^2 = 3.8242

t = 1.96 s

So the mailbag reaches the ground after 1.96 seconds