The film, Mad Max Fury Road, is known for its action scenes and stunts. One of the stunts shown in the film is a classic motorcycle jump. The bad guys are trying to kill Max and his crew so they repeatedly jump over his truck and drop bombs on it. The truck is about 4 meters tall and the jumps are at a 45 degree angle. What is the velocity required to clear the truck?
Truck Height = 4 m
Jump Angle = 45 degrees
Initial Velocity = ???
Time = ???
Acceleration due to Gravity = -9.8 m/s^2
First we need to find the time. The easiest way to do this would be to use symmetry. It should take the same amount of time for the motorcycle to reach the maximum height as it would for him to fall back down to the ground. So with that logic if we use this equation ---> Yf = Yi + Vi,y + 1/2 ay t^2
We should be able to find the time by replacing Yf with 0 m for the ground, Yi with 4 m for the top of the truck, and Vi,y with 0 m/s due to the fact that it is at the height of the jump. The equation should look like this now ---> 0 = 4 + 0 + 1/2 (-9.8) X^2
With the use of some quick Algebra, we find the time for one half of the jump to be 0.9 seconds. Now that we have this knowledge, we can plug the numbers into this equation to find the initial velocity in the Y direction ---> Vf,y = Vi,y + ay t
If you plug in the numbers then it should look like this ---> 4 = X + (-9.8)(0.9)
We should find the value of the initial Y velocity to be 12.82 m/s.
Then we plug those numbers into this equation to find the initial velocity ---> Vy,i = Vi sin(angle)
It should look like this ---> 12.82 = X sin45
Then we get the initial velocity to be 18.13 m/s. The motorcycle has to be going at 18.13 m/s to make the jump over a 4 meter tall truck. However, it is true that the stuntmen probably want a wide gap between themselves and the truck during the trick, so they would probably go a little faster to get a safer distance from the truck. They most likely went at a minimum speed of 20 m/s if I had to guess.