Question

The height of a helicopter above the ground is given by h = 2.65t³, where h is in meters and t is in seconds. After 2.10 s, the helicopter releases a small mailbag. Assume the upward direction is positive and the downward direction is negative.
a.) What is the velocity of the mailbag when it is released?

b.) What maximum height from the ground does the mailbag reach?

c.) What is the velocity of the mailbag when it hits the ground?

d.) How long after its release does the mailbag reach the ground?

e.) Draw position, velocity, and acceleration graphs. These must be turned in on paper.

Answer 1

h = 2.65 t^3 = 2.65 (2.10)^3 = 24.54 meters
a) s = (2.65)(t^3)
velocity = ds/dt = (2.65)(3)(t^2) = (2.65)(3)(4.41) =
velocity when released = + 35.06 m/s
b)

(time at which velocity = 0):
v = at = 35.06 = (9.8m/s^2)t
t = 3.6 seconds of additional travel upwards.

Thus, maximum mailbag height = s:

s = 24.54m + 35.06*3.6 - (0.5)*9.8*3.6^2
s = 87.252 meters
c)First, find the travel time of the falling mailbag:

s = (0.5)at^2

t^2 = 2s/a = 2(87.252)/9.8 =17.81 s^2
t = 4.22 s
velocity = at = (9.8)(4.22) = 41.356 m/s


d) t = 4.22 s (from step c above)