The height of a helicopter above the ground is given by h = 2.55 t^3, where h is
ID: 1423178 • Letter: T
Question
The height of a helicopter above the ground is given by h = 2.55 t^3, where h is in meters and t is in seconds. After 1.60 s, the helicopter releases a small mailbag. Assume the upward direction is positive and the downward direction is negative. What is the velocity of the mailbag when it is released? What maximum height from the ground does the mailbag reach? What is the velocity of the mailbag when it hits the ground? How long after its release does the mailbag reach the ground? Draw position, velocity, and acceleration graphs. These must be turned in on paper.Explanation / Answer
1.The height of the helicopter from the ground is given as a cubic function of time.
h = 2.55t³
After 1.6 s, the height is h= 2.55*(1.6)³ = 10.44m/s
2.Differentiating h, we will get the velocity as,
u = dh/dt = 3*2.55t² = 7.65t²
At time t = 1.6 s, u = 7.65*(1.6)² = 19.58 m/s. So this is your initial velocity!
3.Now we will find t:
You are familiar with the kinematic equation of motion,
S = ut + ½at²
S is the displacement of the body, in this case it would be the distance travelled from the helicopter to the ground, = Y - Yo
Yo = 10.44m, Y = 0
a = -g, downward
Y - Yo = ut+ ½at²
--> 0 - 10.44 = 19.58t -½gt²
--> ½gt² - 19.58t - 10.44=0
--> 4.9t² - 19.58t - 10.44 = 0
So now we have a quadratic in t. Solve it and we will get the time taken by the bag to reach the ground.
t = - b±(b²-4ac)/2a
= 19.58 ± (19.58²-4*4.9*-10.44)/2*4.9
= 19.58 ± 24.25/9.8
= 4.47,-0.47
Hence the time t = 4.47s
Now we are ready to answer the questions,
a) velocity of the mailbag when it isreleased = 19.58m/s
b) Maximum height from the ground the mailbag reaches = 10.44 m
c) v = u + at
= 19.58 - 9.8*4.47
= -24.226 m/s, downwards
d) Time taken by the bag to reach the ground, t = 4.47s
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