A helicopter carrying supplies is traveling eastward, 80.0 m off the ground, at a speed of 40.0 m/s. The supplies are to be drop

Question

4 years ago

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A helicopter carrying supplies is traveling eastward, 80.0 m off the ground, at a speed of 40.0 m/s. The supplies are to be drop

ped from the helicopter to land in the back of a parked pickup truck on a flat road. When the helicopter and the truck are 300 m apart,
(a) How many seconds later should the helicopter crew drop the supplies?
(b) If the pickup truck was traveling westward at 25.0 mi/h from the beginning, then at what time should the helicopter crew drop the supplies?

Physics

1 answer:

dezoksy [38] · Answer 1 · 2021-11-12T12:35:13+03:00

Answer:

a) Helicopter need to drop the supply after 3.46 seconds.

b) Helicopter need to drop the supply after 1.83 seconds.

Explanation:

a) Let us find time of flight of supplies that is vertical motion of supplies,

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 80 m

Substituting,

s = ut + 0.5 at²

80 = 0 x t + 0.5 x 9.81 x t²

t = 4.04 s

Distance traveled by helicopter in 4.04 seconds at 40 m/s = 4.04 x 40 = 161.54 m

Distance helicopter can fly without dropping supply = 300 - 161.54 = 138.46 m

Time taken for this

$t=\frac{138.46}{40}=3.46s$

So helicopter need to drop the supply after 3.46 seconds.

b) Speed of truck = 25 mi/h = 11.11 m/s westward

Speed of helicopter = 40 m/s eastward

Relative velocity, = 40 + 11.11 = 51.11 m/s

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 80 m

Substituting,

s = ut + 0.5 at²

80 = 0 x t + 0.5 x 9.81 x t²

t = 4.04 s

Distance traveled by relative bodies in 4.04 seconds at 51.11 m/s = 4.04 x 51.11 = 206.48 m

Distance helicopter can fly without dropping supply = 300 - 206.48 = 93.52 m

Time taken for this

$t=\frac{93.52}{51.11}=1.83s$

So helicopter need to drop the supply after 1.83 seconds.