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Lynna [10]
3 years ago
6

During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying

in your chaise lounge soaking up the warm Mediterranean sun, a large glob of seagull poop hits you in the face. Since you got an “A” in Physics you are able to estimate the impact velocity at 98.5 m/s. Neglecting air resistance, calculate how high up the seagull was flying when it pooped
Physics
1 answer:
Jlenok [28]3 years ago
8 0

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.

That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:

             Height = (1/2) (acceleration) (time²) .

If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.

                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
                 Height = (1/2) (9.81) (10.04)²

                            =   (4.905 m/s²) x (100.8 sec²)  =  494.43 meters.

As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.

If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component.  That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.   

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a)  please find the attachment

(b) 3.65 m/s^2

c) 2.5 kg

d) 0.617 W

T<weight of the hanging block

Explanation:

a) please find the attachment

(b) Let +x be to the right and +y be upward.

The magnitude of acceleration is the same for the two blocks.  

In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:  

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c) <em>in order to calculate m we will apply newton second law on the hanging   </em>

<em>    block</em>

<em> </em>∑F=ma_y

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<em>the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax</em>

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W=2.5*9.8

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T=14.7/25

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T<weight of the hanging block

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