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According to a newspaper account, a paratrooper survived a training jump from 1200 ft when his parachute failed to open but provided some resistance by flapping unopened in the wind. Allegedly he hit the ground at 100 mi/h after falling for 8 sec. Test the accuracy of this account. dv (Suggestions: Use the model -Pv – g. Using the model, calculate v(8) dt and y(8). Solve v(8) for e⁻⁸ᵖ – 1. From there, find an exact value for p and determine if this will give information that matches the newspaper article. Also, be careful with units!)

User Ken Hung
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Final answer:

The newspaper account's accuracy can be assessed using the differential equation dv/dt = -Pv - g, converting units to metric, and solving for P. An example of deceleration computed using the kinematics equation is the airmen from WWII who decelerated at 486 m/s² when stopped by snow and trees after a fall with an initial speed of 54 m/s over 3 meters.

Step-by-step explanation:

The accuracy of the newspaper account stating a paratrooper hit the ground at 100 mi/h after falling for 8 seconds can be tested using the differential equation dv/dt = -Pv - g, where P is the resistance factor due to the parachute flapping in the wind and g is the acceleration due to gravity. Assuming standard gravity of 9.8 m/s², we must convert the impact speed to metric units (100 mi/h is approximately 44.7 m/s). By integrating the equation and using the boundary conditions, we solve for the resistance factor P that would result in the stated velocity after 8 seconds of free fall.

For an example with known deceleration, consider the World War II airmen who survived falls by decelerating over a distance due to snow drifts and tree branches. If an airman's speed upon impact was 54 m/s (123 mph) and they were stopped over 3.0 meters, we can use the kinematic equation v² = u² + 2as (where u is initial velocity, v is final velocity, a is acceleration, and s is distance) to calculate the deceleration. Initial velocity u = 54 m/s, final velocity v = 0 m/s, and distance s = 3.0 m, leading to a deceleration of 486 m/s².

User Chrispanda
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