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Physics of Hot Rod

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One of the most amazingly idiotic movies ever created is Hot Rod, starring Andy Samberg as stuntman Rod Kimble. In one scene, he attempts to jump a public swimming pool while on his moped. He fails miserably, as the video below illustrates. How could he have successfully jumped the pool? In order to determine this, lets consider the physics behind the situation. Lets say that the ramp is angled at 45 degrees to the horizontal. Lets also say the the the two ramps on the two sides of the pool are at equal heights. The width of the pool is approximately 15 meters. How fast would Rod need to be going while at the top of the ramp in order to clear the pool? We can use kinematics to solve this problem. Let v = the velocity at which he leaves the top of the ramp in order to clear the pool. The time for which he is in the air (t) equals the change in vertical velocity over the acceleration due to gravity [ t = (Vfinal-Vinitial)/g ]. Since the initial velocity in the vertical direction is Vsin45 and the final velocity in the vertical direction is -Vsin45, t can be solved for in terms of V. Therefore, t = (2^(1/2) / g) V . Now lets look at the horizontal plane. t = (2^(1/2) / g) V , acceleration in the x direction equals 0 m/s^2, and his displacement in the x direction equals 15 m. Since x= Vinitial t + 0.5at^2 and a = 0, x = Vinitial t . Since the horizontal component of V is Vcos45, we can solve for V by plugging in 15 meters for x , (2^(1/2) / g)V for t, and Vcos45 for Vinitial. Through this calculation, we find that Rod would need to be traveling at a speed no less than 12.12 m/s at the top of the ramp.

We can also think of the scenario in terms of energy in order to determine how much work his moped will have to do in order to reach this velocity at the top of the ramp. Lets say that the weight of his moped is 275 kg. Lets also assume that friction is negligible, meaning that mechanical energy is conserved in this situation. Where he begins at the top of the hill adjacent to the pool is about 2.0 meters higher than the top of the ramp (y = 2m). He travels about 10 m before reaching the top of the ramp. Also, his bike starts from rest. When he reaches the top of the ramp, all mechanical energy is in the form of kinetic energy. Therefore, an equation to model the conservation of mechanical energy in this scenario could be:

mgy + work done by the moped = 0.5mV^2


We can solve for the work done by the moped through algebraic rearrangement and by plugging in our known values: y = 2 meters and V = 12.12 m/s. Therefore, the work done by the moped is 14882.5 Joules. Since Rod travels a distance of approximately 10 meters before reaching the top of the ramp, and since Work = Fd, the force that his moped would need to apply for him to be able to clear the pool would be 1482.25 Newtons. He might need to buy a more powerful bike in order to be successful in doing this stunt. 


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I love this movie! Great job with the kinematics. I thought it was cool you thought about it in terms of energy too. 

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Seems like a much easier calculation to use energy. What is the distance he could travel going at his original speed?

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