The height of a helicopter above the ground is given by h = 2.55t3, where h is in meters and t is in seconds. At t = 1.60 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

To find out how long after its release the mailbag reaches the ground, we need to determine when the height h is equal to zero.

Given that the height of the helicopter above the ground is given by h = 2.55t^3, we can set up the equation:

2.55t^3 = 0

To solve for t, we can divide both sides of the equation by 2.55:

t^3 = 0 / 2.55

Simplifying, we get:

t^3 = 0

Since any number raised to the power of 3 will be zero if and only if that number is zero, we know that t = 0 is the solution.

Therefore, the mailbag reaches the ground at t = 0 seconds, which means it reaches the ground immediately after it is released.

To find out how long after its release the mailbag reaches the ground, we need to find the time at which the height of the mailbag (h) is equal to zero.

Given that the height of the helicopter above the ground is given by the equation h = 2.55t^3, we can substitute h with zero and solve for t.

0 = 2.55t^3

To solve for t, we can divide both sides of the equation by 2.55:

0 / 2.55 = (2.55t^3) / 2.55

0 = t^3

Since any number cubed is still equal to zero only if the number itself is zero, we can conclude that t = 0.

Therefore, the mailbag reaches the ground at t = 0 seconds.

The vertical velocity is

V = dh/dt = 7.65 t^2 (from differential calculus)
At t = 1.60 s, the height is
h = 10.45 m
and the velocity is
V = 19.58 m/s

Now let t be measured from the time of release. The mailbag reaches the ground when

H = 10.45 + 19.58 t - (g/2)t^2 = 0

Solve for t.