Question

My Notes Notes Ask Your Te O Ask Your Toachar 1 points | Previcun Anewers SerPSETO 2 P50 325t3 where h is in meters and t is in se nts At t·LAO s the he ooter releases a sm all amag H i no The height of a helicopter above the ground is given by h after its release does the mailbag reach the ground? Submit Answer Save Presgress My Notes Ask Your

Answer 1

• H = 3.25t^3
dH/dt = 9.75t^2

at time t = 1.80s, H = 18.954m.
The velocity of the mailbag is u = 31.59m/s. (upward)
Acceleration due to gravity = - 9.8m/s^2
Displacement h = - 18.954 m. (minus since it is below the origin)

From v^2 -u^2 = 2as
v^2 = 2*(-9.8) *(-18.954) +31.59^2
v = - 37.00 m/s taking the minus value since the velocity we want is down ward.

• Distance = average velocity * time
-18.954 = [31.59 + (-37.00)] *t /2
t =7.00 s
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