The height of a helicopter above the ground is given by

NetherCraft 0
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?

Answer

velocity of the helicopter:

dh/dt = v = d(3.25t3)/dt = 3 * 3.25 * t2 =
9.75 t2

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

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h = -0.5 * g t2 + vo 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


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