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# You will prepare and submit a term paper on The Classical Mechanics. Your paper should be a minimum of 1500 words in length.

You will prepare and submit a term paper on The Classical Mechanics. Your paper should be a minimum of 1500 words in length.  .

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Solution: solving the problem would involve designing a setup, as shown in the figure below. The initial conditions are determined, followed by determining the acceleration after the CS is chosen.

The initial conditions:

 . . . . xo = 0. . vox . = . vo cos . . . . yo = 7 ft . . . . . . voy . = . vo sin .

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The acceleration is given by:

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 . . . . . ax = 0 . . . . . . . ay . = . ‑ 32 ft/s2 .

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Inserting these values into the general equations of motion in 2‑dimensions will lead to: . . . . .

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03-4

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 . . x(t) = vo cos  . t . . . . . . . . . . . . . y(t) = ‑ (1/2)(32) t2 + (vo sin ) t + 7 . . . . . . . . . . . . . . . . . vy(t) = . ‑ 32 t + vo . sin  . .

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At t = 1.5 seconds we have: . . . . . . . . x (1.5s) = 30 = vox (1.5) . . . . y(1.5s) = 10 = ‑ 16(1.5)2 + voy(1.5) + 7

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Thus: . . . . . . . . . . . vox = vo cos  . . = 20 ft/sec. . . . . . . . . . . . . . . voy = . vo sin  . . . = . 26 ft/sec.

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The magnitude & direction of the initial velocity is then:

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 . . . vo = . .  . .= . 32.8 ft/sec. tan  . . = (26)/(20)  . . . = 52.4o

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(That is, 52.4o above the horizontal).

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The maximum height above the floor occurs at a time t' when vy(t') = 0. Hence:

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 . . . . . . . . . vy(t') = 0 = ‑ 32 t' + 26 .  . . t' = .866 sec.

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Then . . . . . . y(0.866sec) = ‑ 16 (.866)2 + (26) (.866) + 7 . = . 17.52 ft.

Solution: this is a projectile motion problem. hence solving the problem would involve designing a setup shown in the figure below. The initial conditions are determined followed by determining the acceleration after the CS is chosen.

The initial conditions:

 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x0 = 0. . y0 = 0. . v0x = v0 cos 37. . . v0y = v0 sin 37

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The acceleration is: . . ax = 0. ay = . ‑ 32 ft/sec2.

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03-2

In 2 dimensions, the following is the general equations of motion for constant acceleration:

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 . . x(t) = (1/2) ax t2 + vox t + xo

 . . . y(t) = (1/2) ay t2 + vox t + yo

 . . . . . vx(t) . = . ax t + vox

 . . . . . . vy(t) . = . ay t + voy

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Inserting the known values to get the unknown

 . . x(t) = (48)(4/5) t

 .y(t) = ‑ (1/2)(32) t2 + (48)(3/5) t

vy(t) = . ‑ 32 t + (48)(3/5)

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 . . . . . . . . . y(t') = 0 = ‑ 16 t'2 + (48) (3/5) t'  . t' = 0, or . t' = 1.8 sec.

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Hence the particle will land at x(t') = x (1.8s) = (48)(4/5)(1.8) = . 69 ft from the origin.

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For the second player initial position (t=0) was 100 ft from the origin,

and needs to reach a point 69 ft from the origin

time to be taken is in 1.8 sec for catching the ball. This gives an average velocity of

 . . . . vave = (x2 ‑ x1)/(t2 ‑ t1) = (69 ‑ 100)/(1.8) = ‑ 17 ft/sec.

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The negative sign shows the direction that the player needs to run to, and in this case, it is towards the origin.

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Chapter 6 No. 12

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