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mylen [45]
2 years ago
12

For what purpose the inclined plane is used?​

Physics
1 answer:
Montano1993 [528]2 years ago
3 0

Answer:

Baka makita mo si behati dun

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Show that the optimal launch angle for a projectile subject to gravity is 45o by carrying out the following steps: 6. Write down
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Answer:

sin  2θ = 1    θ=45

Explanation:

They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation

            R = Vo² sin 2θ / g

Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.

We calculate the distance traveled for different angle

          R = vo² Sin (2 15) /9.8

          R = Vo² 0.051 m

In the table are all values ​​in two ways

Angle (θ)                  distance R (x)

 0                 0                     0

15                 0.051 Vo²        0.5 Vo²/g

30                0.088 vo²        0.866   Vo²/g

45                0.102 Vo²        1   Vo²/g

60                0.088 Vo²      0.866   Vo²/g

75                0.051 vo²        0.5   Vo²/g

90                0                     0

See graphic ( R Vs θ)  in the attached ¡, it can be done with any program, for example EXCEL

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A flea jumps straight up to a maximum height of 0.540 m . what is its initial velocity v0 as it leaves the ground?
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For an uniformly accelerated motion, we can write
2ah=v_f^2-v_0^2
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At the maximum height, the velocity is zero, so v_f =0. Therefore we  can solve to find v_0:
v_0 =  \sqrt{2ah}= \sqrt{2(9.81 m/s^2)(0.540 m)}  =3.25 m/s
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