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?

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:

$v^2=v_o{}^2+2a_y\left(h_f-h_i\right)$

Where $v=$ Final speed at the maximum height.

$v_o=$ Initial speed.

Calculation:

From kinematic equation of motions, we get

$v^2=v_o{}^2+2a_y\left(h_f-h_i\right)$

$0^2=v_o{}^2+2\left(-g\right)\left(h_{\max }-0\right)$

$v_o{}^2=-2\left(-g\right)\left(h_{\max }\right)$

$v_o{}^2=2g{h}_{\max }$

We have $h=10m$

$g=9.8m/s^2$

Substituting the values, we get

$v_o{}^2=2\left(9.8m/s^2\right)\left(10m\right)$

$v_o=\sqrt{2\left(9.8m/s^2\right)\left(10m\right)}$

$v_o=14m/s$.

Conclusion:

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

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:

$v^2=v_o{}^2+2a_y\left(h_f-h_i\right)$

Where $v=$ Final speed at the maximum height.

$v_o=$ Initial speed.

Calculation:

From kinematic equation of motions, we get

$v^2=v_o{}^2+2a_y\left(h_f-h_i\right)$

$0^2=v_o{}^2+2\left(-g\right)\left(h_{\max }-0\right)$

$v_o{}^2=-2\left(-g\right)\left(h_{\max }\right)$

$v_o{}^2=2g{h}_{\max }$

We have $h=10m$

$g=9.8m/s^2$

Substituting the values, we get

$v_o{}^2=2\left(9.8m/s^2\right)\left(10m\right)$

$v_o=\sqrt{2\left(9.8m/s^2\right)\left(10m\right)}$

$v_o=14m/s$

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 $14m/s$.