Waht are some good examples of physics in DareDevil?

Answer 1

I'm not quite sure what you mean because everything involves physics all the time. If you want some interesting problems you could do things with Bullseye and how hard he has to throw some of the objects he does to get them stuck in things, or just the projectile motion of anything he throws. Fight scenes always include some crazy jumping around or jumping from building to building that could be turned into some interesting physics problems. That's about all I've got since I haven't seen the movie in a couple years.....

Answer 2

OK Evel Kneevel wants to jump 30 motocycles parked side by side in his own Harley. What's the physics? It's the conservation of energy. Here's how that works.

When Evel's Harley leaves the ramp, it has total energy TE = PE + KE; where PE is the potential energy at the ramp height above ground and KE is the kinetic energy of the bike and Evel on top going V velocity. Assume Evel, if successful, lands on a ramp at the other end of the N = 30 bikes and the ramp is the same height h as the takeoff ramp, which is inclined theta degrees with the horizon.

The conservation of energy law says the TE for the bike and rider are the same anywhere along their trajectory as they pass over the row of parked bikes. Thus, at the maximum height H, TE(H) = PE(H) + KE(H); where max height H = h + S; where S is the distance from the ramps heights to the maximum height H above ground level where h = 0.

What is unique at H is that KE(H) = 1/2 mVx^2; where Vx is the X direction velocity only. Why? Because at H Vy = 0, they are no longer rising and have not yet started back down.

S, in H = h + S, is the critical value because it determines how long Evel's bike will be in the air. This follows because S = 1/2 gt^2 and t = sqrt(2gS), the time going up to H = h + S and back down to h on the receiving ramp at the other end. In other words, the time in the air is T = 2t. Therefore Evel has to stay in the air T seconds to cross all the bikes and land on the far ramp.

Now we about there. If N = 30 bikes side by side cover a distance of D from ramp to ramp, we have D = VxT and Vx = D/T, which is the minimum velocity Evel has to make in a horizontal (along D) direction. And Vx = V cos(theta); where V is the total velocity off the ramp which is inclined at theta degrees with the horizontal.

[NB: Vx really is a minimum, in fact its value will be more because in reality there will be drag forces slowing the bike down. Without knowing the factors that go into drag, we cannot precisely calculate what Vx needs to be. What we do know, however, is that the kinetic energy at the far ramp ke < KE the kinetic energy at the takeoff ramp. And ke is smaller than KE because of that drag which is a work energy WE dissipated as heat. That is ke + WE = KE; so that ke = KE - WE and all three terms vary as V^2.]

And there we have it, given the distance D to jump, the angle and height of the ramp, we can find Vx and consequently V, the velocity Evel's bike and he must be going to make it across the N bikes parked side by side and land on the ramp at the other side. That's a lot of physics and Evel knows just how to solve these problems.

Another example, which I won't detail here, is in flying aerobatics. Here drag, lift, KE, PE, and such all play a part in safely doing what the Blue Angels do routinely.