Ultimate Slip and Slide (cute, but fake)
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Stumble ThisI just loved Popsci.com's analysis of why this video has to be fake:
Now that looks like fun. Of course we intuitively know it's completely fake, and involves the usual videographic sleight of hand, but let's apply some basic physics to the situation to check our intuition.
First of all, the "diver," after being launched off the ramp, travels a horizontal distance of at least 50 meters before landing in the pool. Assuming a 45-degree launch angle, which gives us the maximum trajectory if we neglect air resistance (air resistance will only make matters worse), let's calculate the absolute minimum speed we need to leave the ramp to stick the landing. What we have here is a good old-fashioned projectile!
Our projectile (again neglecting air friction) will have a constant velocity in the horizontal direction. Thus we get a horizontal motion equation as follows:
Δx = v0 cos (45) t
where Δx is the horizontal distance of 50 meters, v0 is the launch velocity, and t is the time of flight.
In the vertical acceleration we have a constant downward acceleration due to gravity of g = 9.8 m/s2. Assuming we are launched from approximately the same height where we land, we get
Δy = 0 = v0 sin (45) t + ½ at2
Combining the two equations, doing a little algebra, and solving for v0 we find that
v0 = 22 m/s or about 50 miles per hour!
Now, are we to believe that the diver is able to get up to such a speed by sliding down that slip-and-slide? Not likely. Being extremely generous by neglecting the very non-negligible force of friction (yes, even on a slip-and-slide), estimating that the diver is at most maybe 10 meters higher than his launch point, then using conservation of energy, we can calculate the maximum possible speed he would have on takeoff. This is going to be a severe overestimate, due to our neglecting friction, but let's see what we get.
According to conservation of energy, we know that the potential energy (mgΔh) that he loses due to the height difference between the start and the launch point will be converted to kinetic energy (½ mv2).
mgΔh = ½ mv2
where v is his launch velocity. We get v = 14 m/s (around 30 miles per hour).
The physics don't lie. It canna' be done, Cap'n!
Now that looks like fun. Of course we intuitively know it's completely fake, and involves the usual videographic sleight of hand, but let's apply some basic physics to the situation to check our intuition.
First of all, the "diver," after being launched off the ramp, travels a horizontal distance of at least 50 meters before landing in the pool. Assuming a 45-degree launch angle, which gives us the maximum trajectory if we neglect air resistance (air resistance will only make matters worse), let's calculate the absolute minimum speed we need to leave the ramp to stick the landing. What we have here is a good old-fashioned projectile!
Our projectile (again neglecting air friction) will have a constant velocity in the horizontal direction. Thus we get a horizontal motion equation as follows:
Δx = v0 cos (45) t
where Δx is the horizontal distance of 50 meters, v0 is the launch velocity, and t is the time of flight.
In the vertical acceleration we have a constant downward acceleration due to gravity of g = 9.8 m/s2. Assuming we are launched from approximately the same height where we land, we get
Δy = 0 = v0 sin (45) t + ½ at2
Combining the two equations, doing a little algebra, and solving for v0 we find that
v0 = 22 m/s or about 50 miles per hour!
Now, are we to believe that the diver is able to get up to such a speed by sliding down that slip-and-slide? Not likely. Being extremely generous by neglecting the very non-negligible force of friction (yes, even on a slip-and-slide), estimating that the diver is at most maybe 10 meters higher than his launch point, then using conservation of energy, we can calculate the maximum possible speed he would have on takeoff. This is going to be a severe overestimate, due to our neglecting friction, but let's see what we get.
According to conservation of energy, we know that the potential energy (mgΔh) that he loses due to the height difference between the start and the launch point will be converted to kinetic energy (½ mv2).
mgΔh = ½ mv2
where v is his launch velocity. We get v = 14 m/s (around 30 miles per hour).
The physics don't lie. It canna' be done, Cap'n!
| 1 month 1 week ago | CH
