Problem I -
Ethan Hunt needs to jump a gap in the bridge. We know he makes the jump, but
the question is, what angle would he need to take off at in order to make the
jump?
Known: We know
that he does make the jump, and after a quick Google search, we found out his
height is 5 feet 7 inches.
Relevant
Quantities: In order to solve for the angle, we need to do trigonometry, but to
get there, we need values for the triangle; more specifically, the velocity in
the x and y axis at takeoff. To solve for those, the values we need are…
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The
size of the gap in the bridge.
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Ethan’s
velocity right before he jumps.
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Forces
acting on him after takeoff, which since we’re ignoring air resistance, is only
gravity.
After analyzing
the jump several times in the movie, I was able to come to the conclusion that
the gap is roughly three and a half “Tom Cruises” wide (or about 19 feet). In
order to find his velocity, I went outside and did several sprints too see what
top speed I could achieve. I’m a runner, so I’m in pretty good shape and after
several tests I averaged 7 meters per second. Granted his character might be in
slightly better shape than I am, we have to consider that he was also wearing
pants and a jacket which would slow him down so were going to assume his
initial velocity is 7 meters per second. And finally, we know gravity is -9.8
meters per second per second.
Since we know
the initial velocity in the X-axis, and we know the distance he has to travel,
we can calculate the time it takes to do so, which comes out to 0.83 seconds.
We can then use this to figure out the time in the Y-Axis because the times
will be identical. We then have to solve for the initial velocity in the Y axis
and we do that by taking the 3 known quantities that we have; time, acceleration due to gravity, and the change
in distance in the Y-Axis which is 0 meters. We know the change in distance in
the Y-Axis is 0 meters because the side of the bridge that he is landing on is
identical to the height of the side that he is taking off on.
Putting those
numbers together in the appropriate Kinematic Equation gives us 4.067 meters
per second for the initial velocity. Since we now know the initial velocities
in both the x and y axis, we can use trigonometry to solve for theta. In the
calculator, you would do “the inverse tan function of the value of the side
opposite of theta, divided by the value of the side adjacent to theta.
Assuming you
entered that properly in the calculator and you’re in degrees mode, not
radians, you should have gotten a value for theta being 30.16 degrees. That is the angle that Ethan Hunt has to jump at to
just clear the gap.
Problem II – At
the end of the film, Ethan Hunt needs to run one mile across town to get to the
location where the “Rabbits Foot” is hiding. The question is, what speed does
he average on his way there/is that speed even humanly possible?
Known: We know
that the “Rabbits Foot” is one mile away because he tells us that in the movie
when he is talking on the phone. If you time the run, starting with when he exits
the window onto the rooftop, to when he arrives at the door of the other place,
it is exactly 100 seconds.
Relevant
Quantities: In order to solve for his average velocity on this run, all we need
to know is the distance he had to travel, and the time it took him to get
there.
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About
1609 meters.
Ø
Exactly
100 seconds.
Knowing that
Velocity is equal to the change in distance divided by the change in time, all
we have to do is plug in for both those values and we get an average velocity
of 16.098 meters per second. That’s
about 36 miles per hour.
The highest
human “foot” speed ever recorded was 27.79 miles per hour (12.4 meters per
second); which was set by Usain Bolt and
recorded during a 100-meter dash in the Olympics. So we can conclude that
either Tom Cruise is the fastest man in the world, or this scene is not
physically, nor humanly possible!
Problem III – Ethan
Hunt leaps off the top of a building running full speed. He is attached to a
wire that when engaged, will swing him from one building to another at which
point we will detach himself and fall down to the roof of the second building.
The question is; is that daring stunt even remotely close to being physically
possible?
Known: We know
that the height of the first building is 226 meters, and the height of the
second building that he needs to land on is only 162 meters. We assume that Tom Cruise can once again reach
his top speed of 7 meters per second when he goes to jump off the building. And
finally, we know that the gap between the two buildings is 47.5 meters.
Relevant
Quantities: There are numerous relevant quantities in this problem because
there are multiple steps that will go into solving whether it’s possible or
not. The relevant quantities include…
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The
height of the first building; 226 meters.
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The
height of the second building; 162 meters.
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The
distance between them; 47.55 meters.
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His initial
velocity in the x- axis when he jumps; 7 meters per second.
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Initial
velocity in the y-axis when he jumps which we can assume is zero considering he
pretty much runs straight right off the top of the building.
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The
time that he is in free fall (before the wire is engaged).
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The
length of the wire.
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How
high off the ground he is when the wire does engage and he starts the swing to
the second building.
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Wind
resistance during his free fall (after a few calculations it came out to an
average of 350 N of force which would reduce his acceleration to -8.1 meters
per second per second.
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The
time he is physically “swinging” on the pendulum which is roughly 12.6 seconds.
Knowing all of
that, you will be able to solve for the distance he travels in the x and
y-axis, which comes out to 45.5 meters and -171.1 meters respectively. Using
the Pythagorean theorem, we can solve for the length of the rope which comes
out to 177.0 meters. Using trig we can solve for the angle of the rope with
respect to vertical which comes out to 15 degrees, and we can deduce that at
the instant that the wire engages, Ethan Hunt is roughly 54.9 meters above the
ground because if he has traveled -171.1 meters vertically, and the building is
originally 226 meters tall (basic subtraction).
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