. In January 2003, an 18-year-old student gained a bit of fame for surviving—with only minor injuries—a remarkable traffic accident. The vehicle he was driving was "clipped" by another one, left the road, and rolled several times. He was thrown upward from the vehicle (he wasn't wearing a seat belt) and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters. (a) Estimate the driver's vertical speed when he was thrown from the vehicle. (b) If he had not landed in the wires, how fast would he have been going when he hit the ground?

Question

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Answer 1

Textbook Question

Chapter 3, Problem 30P

. In January 2003, an 18-year-old student gained a bit of fame for surviving—with only minor injuries—a remarkable traffic accident. The vehicle he was driving was "clipped" by another one, left the road, and rolled several times. He was thrown upward from the vehicle (he wasn't wearing a seat belt) and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.
(a) Estimate the driver's vertical speed when he was thrown from the vehicle.
(b) If he had not landed in the wires, how fast would he have been going when he hit the ground?

Expert Solution

To determine

(a)

The driver’s vertical speed when he was thrown from the vehicle.

Answer to Problem 30P

The speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Explanation of Solution

Given:

In January 2003, an 18 year old student gained a bit of fame for surviving- with only minor injuries-a remarkable traffic accident. The vehicle he was driving was “clipped” by another one, left the road, and rolled several times. He was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s.

Conclusion:

Hence, the speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Expert Solution

To determine

(b)

If the driver had not landed in the wires, how fast would he have been going when he hit the ground?

Answer to Problem 30P

The speed of driver before hitting with the ground is 14m/s.

Explanation of Solution

Given:

Given that a person was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s

Thus, speed of driver before hitting with the ground is also 14m/s.

Conclusion:

Hence, the speed of driver before hitting with the ground is 14m/s.

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A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

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Answer 2

Textbook Question

Chapter 3, Problem 30P

. In January 2003, an 18-year-old student gained a bit of fame for surviving—with only minor injuries—a remarkable traffic accident. The vehicle he was driving was "clipped" by another one, left the road, and rolled several times. He was thrown upward from the vehicle (he wasn't wearing a seat belt) and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.
(a) Estimate the driver's vertical speed when he was thrown from the vehicle.
(b) If he had not landed in the wires, how fast would he have been going when he hit the ground?

Expert Solution

To determine

(a)

The driver’s vertical speed when he was thrown from the vehicle.

Answer to Problem 30P

The speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Explanation of Solution

Given:

In January 2003, an 18 year old student gained a bit of fame for surviving- with only minor injuries-a remarkable traffic accident. The vehicle he was driving was “clipped” by another one, left the road, and rolled several times. He was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s.

Conclusion:

Hence, the speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Expert Solution

To determine

(b)

If the driver had not landed in the wires, how fast would he have been going when he hit the ground?

Answer to Problem 30P

The speed of driver before hitting with the ground is 14m/s.

Explanation of Solution

Given:

Given that a person was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s

Thus, speed of driver before hitting with the ground is also 14m/s.

Conclusion:

Hence, the speed of driver before hitting with the ground is 14m/s.

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Answer 3

Textbook Question

Chapter 3, Problem 30P

. In January 2003, an 18-year-old student gained a bit of fame for surviving—with only minor injuries—a remarkable traffic accident. The vehicle he was driving was "clipped" by another one, left the road, and rolled several times. He was thrown upward from the vehicle (he wasn't wearing a seat belt) and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.
(a) Estimate the driver's vertical speed when he was thrown from the vehicle.
(b) If he had not landed in the wires, how fast would he have been going when he hit the ground?

Expert Solution

To determine

(a)

The driver’s vertical speed when he was thrown from the vehicle.

Answer to Problem 30P

The speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Explanation of Solution

Given:

In January 2003, an 18 year old student gained a bit of fame for surviving- with only minor injuries-a remarkable traffic accident. The vehicle he was driving was “clipped” by another one, left the road, and rolled several times. He was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s.

Conclusion:

Hence, the speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Expert Solution

To determine

(b)

If the driver had not landed in the wires, how fast would he have been going when he hit the ground?

Answer to Problem 30P

The speed of driver before hitting with the ground is 14m/s.

Explanation of Solution

Given:

Given that a person was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s

Thus, speed of driver before hitting with the ground is also 14m/s.

Conclusion:

Hence, the speed of driver before hitting with the ground is 14m/s.

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Answer 4

Textbook Question

Chapter 3, Problem 30P

. In January 2003, an 18-year-old student gained a bit of fame for surviving—with only minor injuries—a remarkable traffic accident. The vehicle he was driving was "clipped" by another one, left the road, and rolled several times. He was thrown upward from the vehicle (he wasn't wearing a seat belt) and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.
(a) Estimate the driver's vertical speed when he was thrown from the vehicle.
(b) If he had not landed in the wires, how fast would he have been going when he hit the ground?

Expert Solution

To determine

(a)

The driver’s vertical speed when he was thrown from the vehicle.

Answer to Problem 30P

The speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Explanation of Solution

Given:

In January 2003, an 18 year old student gained a bit of fame for surviving- with only minor injuries-a remarkable traffic accident. The vehicle he was driving was “clipped” by another one, left the road, and rolled several times. He was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s.

Conclusion:

Hence, the speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Expert Solution

To determine

(b)

If the driver had not landed in the wires, how fast would he have been going when he hit the ground?

Answer to Problem 30P

The speed of driver before hitting with the ground is 14m/s.

Explanation of Solution

Given:

Given that a person was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s

Thus, speed of driver before hitting with the ground is also 14m/s.

Conclusion:

Hence, the speed of driver before hitting with the ground is 14m/s.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The driver’s vertical speed when he was thrown from the vehicle.

Answer to Problem 30P

The speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Explanation of Solution

Given:

In January 2003, an 18 year old student gained a bit of fame for surviving- with only minor injuries-a remarkable traffic accident. The vehicle he was driving was “clipped” by another one, left the road, and rolled several times. He was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s.

Conclusion:

Hence, the speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Expert Solution

To determine

(b)

If the driver had not landed in the wires, how fast would he have been going when he hit the ground?

Answer to Problem 30P

The speed of driver before hitting with the ground is 14m/s.

Explanation of Solution

Given:

Given that a person was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s

Thus, speed of driver before hitting with the ground is also 14m/s.

Conclusion:

Hence, the speed of driver before hitting with the ground is 14m/s.

Answer 8

Expert Solution

To determine

(a)

The driver’s vertical speed when he was thrown from the vehicle.

Answer to Problem 30P

The speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Explanation of Solution

Given:

In January 2003, an 18 year old student gained a bit of fame for surviving- with only minor injuries-a remarkable traffic accident. The vehicle he was driving was “clipped” by another one, left the road, and rolled several times. He was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s.

Conclusion:

Hence, the speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The driver’s vertical speed when he was thrown from the vehicle.

Answer 13

Answer to Problem 30P

The speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Answer 14

Explanation of Solution

Given:

In January 2003, an 18 year old student gained a bit of fame for surviving- with only minor injuries-a remarkable traffic accident. The vehicle he was driving was “clipped” by another one, left the road, and rolled several times. He was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s.

Conclusion:

Hence, the speed of the driver’s body when he was thrown from the vehicle is 14m/s.

Answer 15

Expert Solution

To determine

(b)

If the driver had not landed in the wires, how fast would he have been going when he hit the ground?

Answer to Problem 30P

The speed of driver before hitting with the ground is 14m/s.

Explanation of Solution

Given:

Given that a person was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)

Where v= Final speed at the maximum height.

vo= Initial speed.

Calculation:

From kinematic equation of motions, we get

v2=vo2+2ay(hf−hi)

02=vo2+2(−g)(hmax−0)

vo2=−2(−g)(hmax)

vo2=2ghmax

We have h=10m

g=9.8m/s2

Substituting the values, we get

vo2=2(9.8m/s2)(10m)

vo=2(9.8m/s2)(10m)

vo=14m/s

Thus, speed of driver before hitting with the ground is also 14m/s.

Conclusion:

Hence, the speed of driver before hitting with the ground is 14m/s.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

If the driver had not landed in the wires, how fast would he have been going when he hit the ground?

Answer 20

Answer to Problem 30P

The speed of driver before hitting with the ground is 14m/s.

Answer 21

Explanation of Solution

Given:

Given that a person was thrown upward from the vehicle and ended up dangling from an overhead telephone cable and a ground wire about 8 meters above the ground. Rescuers got him down after 20 minutes. It is estimated that he reached a maximum height of about 10 meters.

Formula used:

Kinematic equation of motion states the following equation:

v2=vo2+2ay(hf−hi)