Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
Question
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Chapter 2, Problem 2.22P

(a)

To determine

The maximum range in vacuum.

(a)

Expert Solution
Check Mark

Answer to Problem 2.22P

The maximum range in vacuum is 1_.

Explanation of Solution

Write the expression for the range of the projectile.

  Rmax=v0g        (I)

Here, v0 is the initial velocity, and g is the acceleration due to gravity.

Conclusion:

Substitute, 1 for v0, and g in equation (I) to find the maximum range.

  Rmax=11=1

Therefore, the maximum range in vacuum is 1_.

(b)

To determine

The range in a given medium at an angle θ=0.75.

(b)

Expert Solution
Check Mark

Answer to Problem 2.22P

The range in a given medium at an angle θ=0.75 is 0.491094_.

Explanation of Solution

Consider the linear drag.

The range can be solved from the following equation.

  vy0+vtervx0R+vterτln(1Rvx0τ)=0        (II)

Here, vter is the terminal velocity, vx0, and vy0 be the horizontal and vertical component of initial velocity.

Conclusion:

Substitute, 1 for v0, 1 for vter, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and π4 for θ in equation (I).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cosθ(vterg))=0(1)sin(π4)+1(1)cos(π4)R+(1)(11)ln(1Rv0cos(π4)(11))=00.7071+10.7071R+ln(1R(0.7071)(1))=02.414R+ln(101.414R)=0        (III)

Solve equation (III) to obtain the value of range, R.

  2.414R+ln(101.414R)=0R=0.491094

Therefore, the range in a given medium at an angle θ=0.75 is 0.491094_.

(c)

To determine

The range in a given medium for some selected values of θ=0.4,0.5,0.6,0.7,0.8.

(c)

Expert Solution
Check Mark

Answer to Problem 2.22P

The range in a given medium at an angle θ=0.4 is 0.462875_, θ=0.5 is 0.499439_, θ=0.6 is 0.513562_, θ=0.7 is 0.508474_, θ=0.8 is 0.486909_.

Explanation of Solution

Use equation (II) to solve range for different values of θ.

Conclusion:

Substitute, 1 for v0, 1 for vter, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.4 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cos(0.4)(vterg))=0(1)sin(0.4)+1(1)cos(0.4)R+(1)(11)ln(1Rv0cos(0.4)(11))=00.3894+10.92106R+ln(1R(0.92106)(1))=01.508R+ln(11.0857R)=0        (IV)

Solve equation (IV) to obtain the value of range, R.

  1.508R+ln(11.0857R)=0R=0.462875

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.5 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cos(0.4)(vterg))=0(1)sin(0.5)+1(1)cos(0.5)R+(1)(11)ln(1Rv0cos(0.5)(11))=00.4794+10.8776R+ln(1R(0.8776)(1))=01.6857R+ln(11.1395R)=0        (V)

Solve equation (V) to obtain the value of range, R.

  1.6857R+ln(11.1395R)=0R=0.499439

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.6 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cos(0.4)(vterg))=0(1)sin(0.6)+1(1)cos(0.6)R+(1)(11)ln(1R(1)cos(0.6)(11))=00.5645+10.8253R+ln(1R(0.8253)(1))=01.8958R+ln(11.2117R)=0        (VI)

Solve equation (VI) to obtain the value of range, R.

  1.8958R+ln(11.2117R)=0R=0.513562

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.7 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cos(0.4)(vterg))=0(1)sin(0.7)+1(1)cos(0.7)R+(1)(11)ln(1R(1)cos(0.7)(11))=00.6442+10.7648R+ln(1R(0.7648)(1))=02.1498R+ln(11.3075R)=0        (VII)

Solve equation (VII) to obtain the value of range, R.

  2.1498R+ln(11.3075R)=0R=0.508474

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.8 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cos(0.4)(vterg))=0(1)sin(0.8)+1(1)cos(0.8)R+(1)(11)ln(1R(1)cos(0.8)(11))=00.7173+10.6967R+ln(1R(0.6967)(1))=02.4649R+ln(11.4353R)=0        (VIII)

Solve equation (VIII) to obtain the value of range, R.

  2.4649R+ln(11.4353R)=0R=0.486909

Therefore, the range in a given medium at an angle θ=0.4 is 0.462875_, θ=0.5 is 0.499439_, θ=0.6 is 0.513562_, θ=0.7 is 0.508474_, θ=0.8 is 0.486909_.

(d)

To determine

To calculate range corresponding to smaller interval until the maximum range is obtained.

(d)

Expert Solution
Check Mark

Answer to Problem 2.22P

The maximum range is 0.514043_ at an angle θ=0.62_.

Explanation of Solution

Use equation (II) to find the range.

Conclusion:

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.62 for θ in equation (II.

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cosθ(vterg))=0(1)sin(0.62)+1(1)cos(0.62)R+(1)(11)ln(1Rv0cos(0.62)(11))=00.5810+10.8139R+ln(1R(0.8139)(1))=01.9425R+ln(11.22865R)=0        (IX)

Solve equation (IX) to obtain the value of range, R.

  1.9425R+ln(11.22865R)=0R=0.514043

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.62 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cosθ(vterg))=0(1)sin(0.62)+1(1)cos(0.62)R+(1)(11)ln(1Rv0cos(0.62)(11))=00.5810+10.8139R+ln(1R(0.8139)(1))=01.9425R+ln(11.22865R)=0        (IX)

Solve equation (IX) to obtain the value of range, R.

1.9425R+ln(11.22865R)=0R=0.514043

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.63 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cosθ(vterg))=0(1)sin(0.63)+1(1)cos(0.63)R+(1)(11)ln(1Rv0cos(0.63)(11))=00.5891+10.8080R+ln(1R(0.8080)(1))=01.9667R+ln(11.2376R)=0        (X)

Solve equation (X) to obtain the value of range, R.

1.9667R+ln(11.2376R)=0R=0.513955

Substitute, 1 for v0, and g, vterg for τ, v0cosθ for vx0, v0sinθ for vy0, and 0.64 for θ in equation (II).

  v0sinθ+vterv0cosθR+vter(vterg)ln(1Rv0cosθ(vterg))=0(1)sin(0.64)+1(1)cos(0.64)R+(1)(11)ln(1Rv0cos(0.64)(11))=00.5972+10.8021R+ln(1R(0.8021)(1))=01.9913R+ln(11.2467R)=0        (XI)

Solve equation (XI) to obtain the value of range, R.

1.9913R+ln(11.2467R)=0R=0.513758

Thus from the above calculations the value of R increases up to θ=0.62 and then decreases.

Therefore, the maximum range is 0.514043_  at an angle θ=0.62_.

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