COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 4, Problem 51P

(a)

To determine

The cannonball reaches the maximum height above the ground.

(a)

Expert Solution
Check Mark

Answer to Problem 51P

The cannonball reaches the maximum height above the ground is 36.09m_.

Explanation of Solution

Write the expression for the horizontal initial velocity of the cannonball,

  vx0=v0cosθ        (I)

Here, vx0 is the horizontal initial velocity of the cannonball, v0 is the initial velocity of the cannon ball and θ is the launch angle.

Write the expression for the vertical initial velocity of the cannonball,

  vy0=v0sinθ        (II)

Here, vy0 is the vertical initial velocity of the cannonball, v0 is the initial velocity of the cannonball and θ is the launch angle.

Write the expression for the maximum height of the cannon ball above the ground,

    hmax=h0+(vyf2vy022g)        (III)

Here, hmax is the maximum height of the cannonball above the ground, h0 is the initial height of the cannonball above the ground, vyf is the final vertical velocity of the cannonball and g is the acceleration due to gravity.

Conclusion:

Substitute 40m/s for v0 and 37 for θ in (II),

  vy0=(40m/s)sin37=24.07m/s

Substitute 24.07m/s and vy0, 0 for vyf , 7.0m for h0 and 9.8m/s2 for g in (III),

  hmax=7.0m+(0(24.07m/s)22(9.8m/s2))=7.0m+29.09m=36.09m

Therefore, the cannonball reaches the maximum height above the ground is 36.09m_.

(b)

To determine

The horizontal range of the cannon ball.

(b)

Expert Solution
Check Mark

Answer to Problem 51P

The horizontal range of the cannonball is 156.94m.

Explanation of Solution

Write the expression for the horizontal range of the cannonball.

    R=v02sin(2θ)g        (IV)

Here, R is the horizontal range of the cannon ball.

Conclusion:

Substitute 37 for θ, 9.8m/s2 for g and 40m/s for v0 in above equation.

    R=(40m/s)2sin(2(37))9.8m/s2=156.94m

Therefore, the horizontal range of the cannonball is 156.94m.

(c)

To determine

The x and y components of the cannonball’s velocity.

(c)

Expert Solution
Check Mark

Answer to Problem 51P

The x and y components of the cannonball’s velocity are 31.95m/s&35.87m/s.

Explanation of Solution

Write the expression for the x component of the cannonball’s velocity.

    vx0=v0cosθ

Here, vx0 is the horizontal initial velocity.

Write the expression for the final vertical velocity of the cannonball.

    vyf2=vy02+2ghmax        (V)

Here, vyf is the final vertical velocity of the cannonball.

Conclusion:

Substitute 40m/s for v0 and 37 for θ in (I) from part (a),

  vx0=(40m/s)cos37=31.95m/s

Substitute 9.8m/s2 for g and 24.07m/s for vy0 and 36.09m for hmax in (V).

    vyf2=(24.07m/s)2+2(9.8m/s2)(36.09m)vyf=1286.7289m/s=35.87m/s

Therefore, the x and y components of the cannonball’s velocity are 31.95m/s&35.87m/s.

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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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