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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:

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

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The acceleration is given by:

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

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

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03-4

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

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

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

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The magnitude & direction of the initial velocity is then:

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&nbsp.&nbsp.&nbsp. vo =&nbsp.&nbsp. &nbsp.&nbsp.=&nbsp. 32.8 ft/sec. tan &nbsp.&nbsp. = (26)/(20) &nbsp.&nbsp.&nbsp. = 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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&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp. vy(t’) = 0 = ‑ 32 t’ + 26&nbsp. &nbsp.&nbsp. t’ = .866 sec.

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Then&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp. y(0.866sec) = ‑ 16 (.866)2 + (26) (.866) + 7&nbsp. =&nbsp. 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:

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

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The acceleration is:&nbsp.&nbsp. ax = 0. ay =&nbsp. ‑ 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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&nbsp.&nbsp. x(t) = (1/2) ax t2 + vox t + xo

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

&nbsp.&nbsp.&nbsp.&nbsp.&nbsp. vx(t)&nbsp. =&nbsp. ax t + vox

&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp. vy(t)&nbsp. =&nbsp. ay t + voy

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Inserting the known values to get the unknown

&nbsp.&nbsp. x(t) = (48)(4/5) t

&nbsp.y(t) = ‑ (1/2)(32) t2 + (48)(3/5) t

vy(t) =&nbsp. ‑ 32 t + (48)(3/5)

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&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp.&nbsp. y(t’) = 0 = ‑ 16 t’2 + (48) (3/5) t’ &nbsp. t’ = 0, or&nbsp. t’ = 1.8 sec.

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Hence the particle will land at x(t’) = x (1.8s) = (48)(4/5)(1.8) =&nbsp. 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

&nbsp.&nbsp.&nbsp.&nbsp. 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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