0% Complete

Solutions Key: The Physics of a Hollywood Stunt

Started:

Student Information

Solutions Key: The Physics of a Hollywood Stunt

Note: All solutions will use a coordinate system where “up” is the positive direction and “down” is the negative direction. The acceleration due to gravity is therefore

$ a_g = -9.80 \, \text{m/s}^2 $.

Segment 1 of 3: The Test Fall (Apprentice Skills)

  1. Sandbag Velocity after 3.50 s

    Starting Formula:

    $ v_f = v_i + a_g t $

    Given:

    $ v_i = 0 \, \text{m/s}, \quad t = 3.50 \, \text{s}, \quad a_g = -9.80 \, \text{m/s}^2 $

    Solve:

    $ v_f = 0 \, \frac{\text{m}}{\text{s}} + \left(-9.80 \, \frac{\text{m}}{\text{s}^2}\right)(3.50 \, \text{s}) $

    $ v_f = \left(-9.80 \, \frac{\text{m}}{\cancel{\text{s}^2}}\right)(3.50 \, \cancel{\text{s}}) = -34.3 \, \frac{\text{m}}{\text{s}} $

    $ v_f = -34.3 \, \text{m/s} $

    The velocity is 34.3 m/s downward.

  2. Building Height from Impact Velocity

    Starting Formula:

    $ v_f^2 = v_i^2 + 2a_g \Delta y $

    Rearrange:

    $ \Delta y = \frac{v_f^2 - v_i^2}{2a_g} $

    Given:

    $ v_i = 0 \, \text{m/s}, \quad v_f = -41.65 \, \text{m/s}, \quad a_g = -9.80 \, \text{m/s}^2 $

    Solve:

    $ \Delta y = \frac{\left(-41.65 \, \frac{\text{m}}{\text{s}}\right)^2 - \left(0 \, \frac{\text{m}}{\text{s}}\right)^2}{2\left(-9.80 \, \frac{\text{m}}{\text{s}^2}\right)} $

    $ \Delta y = \frac{1734.7 \, \frac{\text{m}^2}{\text{s}^2}}{-19.6 \, \frac{\text{m}}{\text{s}^2}} = \frac{1734.7 \, \frac{\text{m}^2}{\cancel{\text{s}^2}}}{-19.6 \, \frac{\text{m}}{\cancel{\text{s}^2}}} = -88.5 \, \text{m} $

    The height of the building is the magnitude of the displacement, 88.5 m.

  3. Final Velocity when Thrown Downward

    Starting Formula:

    $ v_f^2 = v_i^2 + 2a_g \Delta y $

    Rearrange:

    $ v_f = \sqrt{v_i^2 + 2a_g \Delta y} $

    Given:

    $ v_i = -10.0 \, \text{m/s}, \quad \Delta y = -75.0 \, \text{m}, \quad a_g = -9.80 \, \text{m/s}^2 $

    Solve:

    $ v_f^2 = \left(-10.0 \, \frac{\text{m}}{\text{s}}\right)^2 + 2\left(-9.80 \, \frac{\text{m}}{\text{s}^2}\right)(-75.0 \, \text{m}) $

    $ v_f^2 = 100 \, \frac{\text{m}^2}{\text{s}^2} + \left(-19.6 \, \frac{\text{m}}{\text{s}^2}\right)(-75.0 \, \text{m}) $

    $ v_f^2 = 100 \, \frac{\text{m}^2}{\text{s}^2} + 1470 \, \frac{\cancel{\text{m}} \cdot \text{m}}{\text{s}^2} = 1570 \, \frac{\text{m}^2}{\text{s}^2} $

    $ v_f = \sqrt{1570 \, \frac{\text{m}^2}{\text{s}^2}} = -39.6 \, \frac{\text{m}}{\text{s}} $

    $ v_f = -39.6 \, \text{m/s} $

    The final velocity is 39.6 m/s downward. (We choose the negative root as the object is moving downward).

C. Metacognitive Prompt Response

In problem #3, the non-zero initial velocity ($ v_i \neq 0 $) meant that I could no longer use simplified versions of the kinematic equations (like $ \Delta y = \frac{1}{2} a_g t^2 $). I had to use a full kinematic equation that explicitly included a term for $ v_i $. The time-independent equation ($ v_f^2 = v_i^2 + 2a_g \Delta y $) was the most direct choice because time was not given or requested. It changed the algebra slightly but not the fundamental approach.

Segment 2 of 3: The Stunt Performer’s Jump (Journeyman Synthesis)

  1. Velocity at Impact with Airbag

    Starting Formula:

    $ v_f^2 = v_i^2 + 2a_g \Delta y $

    Given:

    $ v_i = 0 \, \text{m/s}, \quad \Delta y = -60.0 \, \text{m}, \quad a_g = -9.80 \, \text{m/s}^2 $

    Solve:

    $ v_f^2 = 0 \, \frac{\text{m}}{\text{s}} + 2\left(-9.80 \, \frac{\text{m}}{\text{s}^2}\right)(-60.0 \, \text{m}) $

    $ v_f^2 = \left(-19.6 \, \frac{\text{m}}{\text{s}^2}\right)(-60.0 \, \text{m}) = 1176 \, \frac{\cancel{\text{m}} \cdot \text{m}}{\text{s}^2} = 1176 \, \frac{\text{m}^2}{\text{s}^2} $

    $ v_f = \sqrt{1176 \, \frac{\text{m}^2}{\text{s}^2}} = -34.3 \, \frac{\text{m}}{\text{s}} $

    $ v_f = -34.3 \, \text{m/s} $

    The performer’s velocity is 34.3 m/s downward upon touching the airbag.

  2. Acceleration inside the Airbag

    This is a new phase of motion. The initial velocity here is the final velocity from the previous part.

    Starting Formula:

    $ v_f^2 = v_i^2 + 2a \Delta y $

    Rearrange:

    $ a = \frac{v_f^2 - v_i^2}{2\Delta y} $

    Given:

    $ v_i = -34.3 \, \text{m/s}, \quad v_f = 0 \, \text{m/s}, \quad \Delta y = -4.50 \, \text{m} $

    Solve:

    $ a = \frac{\left(0 \, \frac{\text{m}}{\text{s}}\right)^2 - \left(-34.3 \, \frac{\text{m}}{\text{s}}\right)^2}{2(-4.50 \, \text{m})} $

    $ a = \frac{0 - 1176.5 \, \frac{\text{m}^2}{\text{s}^2}}{-9.00 \, \text{m}} = \frac{-1176.5 \, \frac{\text{m}^2}{\text{s}^2}}{-9.00 \, \text{m}} = +131 \, \frac{\cancel{\text{m}} \cdot \text{m}}{\text{s}^2 \cdot \cancel{\text{m}}} $

    $ a = +131 \, \text{m/s}^2 $

    The acceleration is $ 131 \, \text{m/s}^2 $ upward.

  3. Time to Stop in Airbag

    Starting Formula:

    $ v_f = v_i + at $

    Rearrange:

    $ t = \frac{v_f - v_i}{a} $

    Given:

    $ v_i = -34.3 \, \text{m/s}, \quad v_f = 0 \, \text{m/s}, \quad a = +131 \, \text{m/s}^2 $

    Solve:

    $ t = \frac{0 \, \frac{\text{m}}{\text{s}} - \left(-34.3 \, \frac{\text{m}}{\text{s}}\right)}{131 \, \frac{\text{m}}{\text{s}^2}} $

    $ t = \frac{34.3 \, \frac{\text{m}}{\text{s}}}{131 \, \frac{\text{m}}{\text{s}^2}} = \frac{34.3 \, \frac{\cancel{\text{m}}}{\text{s}}}{131 \, \frac{\cancel{\text{m}}}{\text{s}^2}} = \frac{34.3 \, \text{s}^{\cancel{2}}}{131 \, \cancel{\text{s}}} $

    $ t = 0.262 \, \text{s} $

    It takes 0.262 s for the airbag to stop the performer.

C. Cross-Cutting Lens Response

The acceleration during free-fall is $ -9.80 \, \text{m/s}^2 $, while the acceleration provided by the airbag is $ +131 \, \text{m/s}^2 $. The airbag’s acceleration is over 13 times larger in magnitude. This massive acceleration is required to produce a very large change in velocity (from -34.3 m/s to 0 m/s) in a very short distance. The “effect” of an overly large acceleration would be a dangerously large force on the stunt performer, potentially causing severe injury. The goal of the airbag is to reduce the stopping acceleration (and force) by increasing the stopping distance and time.

Segment 3 of 3: The Mid-Air Rescue (Master’s Analysis)

  1. Position Equations

    Let $ t $ be the time since the artifact was dropped. The ground is at $ y = 0 $.

    Artifact: Starts at

    $ y_0 = 150 \, \text{m} $ with $ v_0 = 0 \, \text{m/s} $.

    $ y_{\text{artifact}}(t) = y_0 + v_0 t + \frac{1}{2} a_g t^2 $

    $ y_{\text{artifact}}(t) = 150 \, \text{m} + 0 + \frac{1}{2}\left(-9.80 \, \frac{\text{m}}{\text{s}^2}\right)t^2 $

    $ y_{\text{artifact}}(t) = 150 \, \text{m} - \left(4.90 \, \frac{\text{m}}{\text{s}^2}\right)t^2 $

    Hero: Starts at

    $ y_0 = 130 \, \text{m} $ with $ v_0 = 25.0 \, \text{m/s} $, but their clock starts at $ t = 1.50 \, \text{s} $. We use $ (t - 1.50) $ for the hero’s time.

    $ y_{\text{hero}}(t) = y_0 + v_0 (t - 1.50) + \frac{1}{2} a_g (t - 1.50)^2 $

    $ y_{\text{hero}}(t) = 130 \, \text{m} + \left(25.0 \, \frac{\text{m}}{\text{s}}\right)(t - 1.50 \, \text{s}) - \left(4.90 \, \frac{\text{m}}{\text{s}^2}\right)(t - 1.50 \, \text{s})^2 $

  2. Time of Intercept

    Set the positions equal:

    $ y_{\text{artifact}}(t) = y_{\text{hero}}(t) $.

    $ 150 - 4.90t^2 = 130 + 25.0(t - 1.50) - 4.90(t - 1.50)^2 $

    Expand the squared term for the hero:

    $ (t - 1.50)^2 = t^2 - 3.00t + 2.25 $

    $ 150 - 4.90t^2 = 130 + 25.0t - 37.5 - 4.90(t^2 - 3.00t + 2.25) $

    $ 150 - 4.90t^2 = 130 + 25.0t - 37.5 - 4.90t^2 + 14.7t - 11.025 $

    The $ -4.90t^2 $ terms cancel on both sides. Combine like terms.

    $ 150 = 81.475 + 39.7t $

    $ 68.525 = 39.7t $

    $ t = \frac{68.525}{39.7} = 1.73 \, \text{s} $

    They meet 1.73 s after the artifact is dropped.

  3. Altitude of Catch

    Use the time from the previous part in the simpler artifact equation.

    $ y_{\text{catch}} = 150 \, \text{m} - \left(4.90 \, \frac{\text{m}}{\text{s}^2}\right)(1.73 \, \text{s})^2 $

    $ y_{\text{catch}} = 150 \, \text{m} - \left(4.90 \, \frac{\text{m}}{\cancel{\text{s}^2}}\right)(2.99 \, \cancel{\text{s}^2}) $

    $ y_{\text{catch}} = 150 \, \text{m} - 14.7 \, \text{m} = 135 \, \text{m} $

    The catch is made at an altitude of 135 m.

C. Qualitative Critique & Extension Response

If the hero waited too long (e.g., 3.00 s), the artifact would have more time to fall and would be moving much faster downward. The hero, jumping with the same upward velocity, might not be able to reach the artifact’s new, lower position in the air. At some point, the artifact will be below the highest point the hero can possibly reach.

To determine the absolute “last second” to jump, you would need to find the conditions where their paths intersect at exactly one point, which would be at the peak of the hero’s trajectory. You would first calculate the time it takes the hero to reach their peak ($ v_f = 0 \, \text{m/s} $) and the height of that peak. Then, you would calculate how long it takes the artifact to fall to that same height. The difference between these two times is the maximum possible delay for the hero to still make the catch. If he waits a moment longer, the artifact will always be below the highest point he can reach.