Wednesday, December 2, 2015

The Physics Behind The Hulk's Jump

How does the concrete of the road not crack? That is the question I asked when I saw the Hulk make his high jump shown at 1:53 minutes in the video attached below (Hulk smash scenes in the Avengers movie). In this scene of the movie, the avengers were in the middle of the fight and Hulk makes a pretty high jump to punch down a robot located on a building approximately 120 meters above ground level. As this enormous creature jumps that high, I expected the concrete road to crack due to the tremendous amount of force applied to it.

Rather, the concrete road seems intact as the Hulk makes that giant jump to get to the robot and smash it. then I started to look at how much force the Hulk exerted on the ground as he makes  the jump.

Assuming the center of mass of the Hulk is located at his waist, as he crouches and makes the jump his center of mass moves a certain distance (ΔS) which will give the Hulk his speed as he leaves the ground. so the velocity with which he leaves the ground (V) may be given as follows:

V^2=Vo^2+2a(ΔS) ------- (1)

Since the initial velocity (Vo) is zero and the acceleration can be given in terms of the force the Hulk applies on the ground and his mass, the above equation may be rewritten as:

V^2=2(F/M)(ΔS) -------- (2)

And assuming his jump comes to a halt as he gets to the robot at a height of 120 meters (i.e. his final velocity at 120 m is 0 m/s), the velocity at which he leaves the ground can be also given by:

V^2=2gh -------- (3)

By setting equations (2) and (3) to be equal, we can find the force (F). Assuming ΔS to be 1.2 m (since Hulk is about 7 foot and 6 inches [about 2.3 m]) and the Hulk's mass is 1220 lbs (554.55 Kg), we can find the force that the Hulk exerts on the ground.

2(F/M) (ΔS) = 2gh
F = (2.g.h.M)/(2.ΔS)
F = (9.8 m/s^2*120 m*554.55 Kg)
F = 543,459 N

From this ,the pressure exerted on the road during the jump was calculated to see if the concrete will break or not. To do this, the area of the Hulk's feet should be found. By scaling up the size of the foot from what I see on my laptop screen, I found the dimension of a foot to be roughly about 0.44 m by 0.16 m which makes the combined area of the two feet about 0.14 m^2. Therefore, the pressure exerted by the foot of the Hulk is:

P = F/A
P = 543,459 N / 0.14 m^2
P = 3.882 MPa

According to the Engineering Toolbox, the ultimate compression strength of concrete ranges from 20 MPa - 40 MPa. This tells us that the concrete road, although it seems otherwise, can be fully support Hulk's jump without breaking.

The following web pages were used to get facts for this post:
• http://www.engineeringtoolbox.com/concrete-properties-d_1223.html
• http://marvel.com/characters/25/hulk