The height of a helicopter above the ground is given by h = 3.25t3,

where h is in meters and t is in seconds. At t = 2.10 s, the

helicopter releases a small mailbag. How long after its release

does the mailbag reach the ground?

where h is in meters and t is in seconds. At t = 2.10 s, the

helicopter releases a small mailbag. How long after its release

does the mailbag reach the ground?

## Answer

velocity of the helicopter:

dh/dt = v = d(3.25t^{3})/dt = 3 * 3.25 * t^{2} =

9.75 t^{2}

velocity of the helicopter t=2.10s :

v = 9.75 * (2.10 * 2.10) = 42.998 m/s

Thereore the initial velocity of the mailbag is 42.998 m/s

initial position of the mailbag = h = 3.25*2.10*2.10*2.10 =

30.098 m

h = -0.5 * g t^{2} + v_{o} t

-30.098 = -0.5 * 9.8 * t^2 + 42.998 * t

==> t = 9.43

s

h = 3.25 t^3

=>u = 3.25*3*t^2

At t = 2.1 s, u = 3.25*3*(2.1^2) = 43 m/s

h = 3.25*(2.1^3) = 30.09825 m

v^2 = u^2+2gh = (43^2)+(2*9.81*30.09825) => v = 49.3916

m/s

v = u+at

=>t = (v-u)/9.81 = (49.39157484-43)/9.81 = 0.65154 s