Question

The height of a helicopter above the ground is given by h = 2.95t³, where h is in meters and t is in seconds. At t = 2.05 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer 1

Gravitational acceleration = g = -9.81 m/s²

Height of the helicopter is given by,

h = 2.95t³

Speed of the helicopter can be found by differentiating this equation with respect to time,

V = 3 x (2.95t²)

V = 8.85t²

At t=2.05 sec the helicopter releases a mailbag.

Height of the mailbag when it is released = H₁

H₁ = (2.95)(2.05)³

H₁ = 25.414 m

Velocity of the mailbag when it is released = V₁

V₁ = (8.85)(2.05)²

V₁ = 37.192 m/s

Time taken by the mailbag to reach the ground = T

When the mailbag reaches the ground it's displacement is in the downward direction therefore it is negative.

-H₁ = V₁T + gT²/2

-25.414 = (37.192)T + (-9.81)T²/2

4.905T² - 37.192T - 25.414 = 0

T = 8.213 sec or -0.63 sec

Time cannot be negative.

T = 8.213 sec

Time taken by the mailbag to reach the ground = 8.213 sec