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Gemiola [76]
3 years ago
7

An airplane is flying horizontally with a speed of 103 km/hr (278 m/s) when it drops a payload. The payload hits the ground 30 s

later. (Neglect air drag and the curvature of the Earth. Take g = 10 m/s².)
At what altitude H is the airplane flying?
Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

H = 4500 m

Explanation:

  • Once dropped, the payload moves along a trajectory, that can be decomposed along two directions independent each other.
  • Just by convenience, we choose these directions to be coincident with the horizontal (-x) and vertical (y) axes.
  • As both movements are independent each other due to both are perpendicular, in the vertical direction, the initial speed is 0.
  • So, in order  to find the vertical displacement at any point in time, we can use the following kinematic equation, where a=-g., and H = -Δy.

        H = \frac{1}{2}*g*t^{2} = \frac{1}{2} * 10 m/s2*(30s)^{2}   = 4500 m

  • The airpane was flying at a 4500 m altitude.
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Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum p_{i} before the collision must be equal to the final momentum p_{f} after the collision:

p_{i}=p_{f} (1)

Being:

p_{i}=MV_{i}+mU_{i}

p_{f}=(M+m) V

Where:

M=480 g \frac{1 kg}{1000 g}=0.48 kg the mas of the peregrine falcon

V_{i}=45 m/s the initial speed of the falcon

m=240 g \frac{1 kg}{1000 g}=0.24 kg is the mass of the pigeon

U_{i}=0 m/s the initial speed of the pigeon (at rest)

V the final speed of the system falcon-pigeon

Then:

MV_{i}+mU_{i}=(M+m) V (2)

Finding V:

V=\frac{MV_{i}}{M+m} (3)

V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg} (4)

V=30 m/s (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

F=\frac{\Delta p}{\Delta t} (6)

Where:

F is the force exerted on the pigeon

\Delta t=0.015 s is the time

\Delta p is the pigeon's change in momentum

Then:

\Delta p=p_{f}-p_{i}=mV-mU_{i} (7)

\Delta p=mV (8) Since U_{i}=0

Substituting (8) in (6):

F=\frac{mV}{\Delta t} (9)

F=\frac{(0.24 kg)(30 m/s)}{0.015 s} (10)

Finally:

F=480 N

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Answer:

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Explanation:

It is given that,

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We need to find the thrust and the force. The mass of the liquid displaced is given by :

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<h2>Speed with which it return to its initial level is 100 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

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Final velocity, v = ?

Displacement, s = 0 m  

Substituting  

v² = u² + 2as

v² = 100² + 2 x -9.81 x 0

v² = 100²

v = ±100 m/s

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