COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 4, Problem 47P

(a)

To determine

Sketch the graphs of x,y, vx and vy as functions of time.

(a)

Expert Solution
Check Mark

Answer to Problem 47P

Figure 1, 2, 3, 4 shows the graphs of x,y, vx and vy as functions of time.

Explanation of Solution

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 4, Problem 47P

Figure 1

Conclusion:

Therefore, Figure 1, 2, 3, 4 shows the graphs of x,y, vx and vy as functions of time.

(b)

To determine

The initial velocity of the stone.

(b)

Expert Solution
Check Mark

Answer to Problem 47P

The initial velocity of the stone is 27.6m/s_ at 25° above the horizontal.

Explanation of Solution

Write the expression for the initial horizontal component of velocity of the stone.

    vix=vicos25°        (I)

Here, vix is the initial horizontal component of velocity of the stone, vi is the initial velocity.

Write the expression for the initial vertical component of velocity of the stone.

    viy=visin25°        (II)

Write the expression for the equation of motion.

    x=vxt+12axt2        (III)

Conclusion:

Substitute 105m for x, 0 for vx, 4.2s for t  in equation (III) to find x.

    105m=vix(4.2s)+12(0m/s2)(4.2s)2vix=105m4.2s=25m/s

Substitute 25m/s for vix in equation (I) to find vi.

    vi=vixcos25°=25m/scos25°=27.6m

Therefore, the initial velocity of the stone is 27.6m/s_ at 25° above the horizontal.

(c)

To determine

The initial height h.

(c)

Expert Solution
Check Mark

Answer to Problem 47P

The initial height h is 37.5m_

Explanation of Solution

Let the initial height of the stone is h. Then at the end of the journey the vertical displacement of the stone is y=h.

The vertical component of initial velocity is ,

    viy=(27.6m/s)sin25°=11.664m/s

Write the expression for the equation of motion.

    y=voyt+12ayt2        (IV)

Conclusion:

At the end of the journey the vertical displacement of the stone is y=h.substitute 11.664m/s for viy , 4.2s for t, 9.8m/s2 for ay in equation (IV) to find h.

    h=(11.664m/s)(4.2s)+12(9.8m/s2)(4.2s)2=(48.988m86.436m)=37.448m

Therefore, the initial height h is 37.5m_

(d)

To determine

The maximum height H that reached by the stone.

(d)

Expert Solution
Check Mark

Answer to Problem 47P

The maximum height H that reached by the stone is 44.44m_ above the ground.

Explanation of Solution

Write the expression relating equation of motion.

    vfy2=viy2+2ayy        (V)

Conclusion:

Substitute 0 for vfy, 11.664m/s for viy and 9.8m/s2 for ay in equation (V) to find y.

    (0m/s)2=(11.64m/s)2+2(9.8m/s2)yy=(11.64m/s)22(9.8m/s2)=6.94m

The maximum height reached by the stone is ,

    H=h+y=6.94m+37.5m=44.44m

Therefore, The maximum height H that reached by the stone is 44.44m_ above the ground.

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Chapter 4 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 4.4 - Prob. 4.8PPCh. 4.5 - Prob. 4.5ACPCh. 4.5 - Prob. 4.10PPCh. 4.5 - Prob. 4.5BCPCh. 4.6 - Prob. 4.11PPCh. 4.6 - Prob. 4.6CPCh. 4 - Prob. 1CQCh. 4 - Prob. 2CQCh. 4 - Prob. 3CQCh. 4 - Prob. 4CQCh. 4 - Prob. 5CQCh. 4 - Prob. 6CQCh. 4 - Prob. 7CQCh. 4 - Prob. 8CQCh. 4 - Prob. 9CQCh. 4 - Prob. 10CQCh. 4 - Prob. 11CQCh. 4 - Prob. 12CQCh. 4 - Prob. 13CQCh. 4 - Prob. 14CQCh. 4 - Prob. 15CQCh. 4 - Prob. 16CQCh. 4 - Prob. 17CQCh. 4 - Prob. 18CQCh. 4 - Prob. 19CQCh. 4 - Prob. 1MCQCh. 4 - Prob. 2MCQCh. 4 - Prob. 3MCQCh. 4 - Prob. 4MCQCh. 4 - Prob. 5MCQCh. 4 - Prob. 6MCQCh. 4 - Prob. 7MCQCh. 4 - Prob. 8MCQCh. 4 - Prob. 9MCQCh. 4 - Prob. 10MCQCh. 4 - Prob. 11MCQCh. 4 - Prob. 12MCQCh. 4 - Prob. 13MCQCh. 4 - Prob. 14MCQCh. 4 - Prob. 15MCQCh. 4 - Prob. 16MCQCh. 4 - Prob. 17MCQCh. 4 - Prob. 18MCQCh. 4 - Prob. 19MCQCh. 4 - Prob. 20MCQCh. 4 - Prob. 21MCQCh. 4 - Prob. 1PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69PCh. 4 - Prob. 70PCh. 4 - Prob. 71PCh. 4 - Prob. 72PCh. 4 - Prob. 73PCh. 4 - Prob. 74PCh. 4 - Prob. 75PCh. 4 - Prob. 76PCh. 4 - Prob. 77PCh. 4 - Prob. 78PCh. 4 - Prob. 79PCh. 4 - Prob. 80PCh. 4 - Prob. 81PCh. 4 - Prob. 82PCh. 4 - Prob. 83PCh. 4 - Prob. 84PCh. 4 - Prob. 85PCh. 4 - Prob. 86PCh. 4 - Prob. 87PCh. 4 - Prob. 88PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92PCh. 4 - Prob. 93PCh. 4 - Prob. 94PCh. 4 - Prob. 95PCh. 4 - Prob. 97PCh. 4 - Prob. 98PCh. 4 - Prob. 99PCh. 4 - Prob. 100PCh. 4 - Prob. 101PCh. 4 - Prob. 102PCh. 4 - Prob. 103PCh. 4 - Prob. 104PCh. 4 - Prob. 105P
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