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

7.96 s

Step-by-step explanation:

Given the following :

h = 3t³

After t seconds, that is, t = 2

h = 3(2)³ = 3 × 8 = 24 m

Rate of change, Velocity = dh/dt

dh/dt = 3*3t^(3 - 1) = 3*3t² = 9t²

At t = 2 s ; initial velocity = 9(2)² = 9 × 4 = 36 m/s

From the 3rd equation of motion:

v² = u² + 2gh

Where g = acceleration due to gravity 9.8m/s

upward motion, g = - ve

Displacement = - 24

v² = 36² + 2(-9.8)(-24)

v² = 1296 + 470.4

v² = 1766.4

v = sqrt(1766.4)

= 42.03 m/s

Since final velocity is in downward direction : take it as negative.

Recall, :

Velocity = Displacement / time

Displacement = Velocity × time

Average Velocity = (Initial + final velocity) / 2

Displacement = - 24 m

-24 = [(36 + (-42.03)) /2] × t

-48 = - 6.03t

t = 48 / 6.03

t = 7.96s

answer:

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step-by-step explanation: