The height of a helicopter above the ground is given by h = 3.20t3, where h is i

Question

The height of a helicopter above the ground is given by h = 3.20t3, where h is in meters and t is in seconds. After 2.20 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?

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

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

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

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please provide how you did your work and give your answer for each

Explanation / Answer

part a:

height=h=3.2*t^3

vertical veloicty=dh/dt=3.2*3*t^2=9.6*t^2

at t=2.2 seconds, vertical velocity=46.464 m/s

so veloicty of mailbag when released is 46.464 m/s

part b:

at t=2.2, height of the mailbag when released=3.2*2.2^3=34.0736 m

initial vertical velocity at that point=46.464 m/s

vertical acceleration=-9.8 m/s^2

it will reach maximum height when its speed becomes 0.

using the formula:

final speed^2-initial speed^2=2*acceleration*distance

==>0^2-46.464^2=-2*9.8*distance

==>distance=110.148 m

hence total maximum height from ground=34.0736+110.148=144.2216 m

time taken to reach maximum height from release=(final speed-initial speed)/acceleration

=(0-46.464)/(-9.8)=4.74122 seconds....(1)

part c:

the mailbag will start at veloicty 0 , from height 144.2216 m

vertical acceleration =9.8 m/s^2(positive as it is along the direction of motion)

using the formula:

final speed^2-initial speed^2=2*acceleration*distance

==>final speed^2-0=2*9.8*144.2216

==>final speed=sqrt(2*9.8*144.2216)=53.167 m/s

hence the mailbag will hit ground with 53.167 m/s

time taken to reach ground from maximum height=(final speed-initial speed)/acceleration

=(53.167-0)/9.8=5.4252 seconds....(2)

part d:

total time taken after release to reach ground=time obtained in equation 1+ time obtained in equation 2

=4.74122+5.4252=10.16642 seconds

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