Question

The height of a helicopter above the ground is given by h = 3.15t3, where h is in meters and t is in seconds. At t = 2.45 s, the helicopter releases a small mailbag.How long after its release does the mailbag reach the ground?

THE ANSWER ISNT 3.07315806

Answer 1

H = 3t^3
dH/dt = 9t^2

at time t = 2.45s, H = 44.12m.
The velocity of the mailbag is u = 54.02m/s. (upward)
Acceleration due to gravity = - 9.8m/s^2
Displacement h = -44.12 m. (minus since it is below the origin)

From v^2 -u^2 = 2as
v^2 = 2*(-9.8) *(-44.12) +54.02^2
v = - 61.50m/s taking the minus value since the velocity we want is down ward.

• Distance = average velocity * time
-44.12 = [54.02 + (-61.50)] *t /2
t =11.80 s