The Physics Behind Surviving a Jump Off an 8-Story Building into Water

Physics Blog 2

Recently, a video was posted on the internet showing a man jumping off an 8-story building into Newport Harbor, California. In the video we see that the man needs to clear two docks in order to safely land in the water buy jumping out a certain distance. As you can infer, this physics question is a real-life example of what we have done in class.

This real-life example represents projectile motion with the addition of the use of a kinematic equation. The man needed to jump over two docks at the base of the building in order to safely land into the water. The following image is a diagram of the persons path of motion in the jump.

(Diagram: Rhett Allain)

The two motions that appear in this circumstance are horizontal motion (x) and vertical motion (y) as a one-dimensional kinematic problem. The gravitation force pulling the man down as he jumps off the building is 9.8 m/s^2 or 32 ft/s^2. The final y value, the water level, is equal to 0, and the initial y value is 129 (129 ft.). The initial y velocity of the man is 0 since he jumped out and not upward. The following kinematic equation can be used to represent the vertical motion with a constant acceleration:

Next, you plug in the values for the starting and ending position, and the initial y velocity to solve for the time.

The following expression can be used for the horizontal (x) motion. There are no horizontal forces on the man, therefor the acceleration here is 0.

Since the acceleration is 0 and we are trying to solve for the starting x value, the following equation can be used to represent the magnitude of the initial velocity vector.

Next, substitute the expression for time, making the horizontal distance traveled be represented in this formula:

Based upon this article and the authors research (Google Maps), the horizontal distance from the initial point to the ending point is approximately 7.4 meters. The height of the building is 129 feet or 39.3 meters, therefor these values can be plugged into the equation the calculate the man's launch speed.

The launch speed comes out to be 2.6 meters per second, or 5.8 miles per hour.

In order to calculate the total air time that the man experienced, the following equation is used.

39.3 = 1/2(9.8)t^2 + 2.6t

39.3 = 4.9t^2 + 2.6t

0 = 4.9t^2 + 2.6t - 39.3

solve for t

t = -3.11, 2.57

Time cannot be negative, therefor the airtime is approximately 2.57 seconds. If you start the video from the time that the man jumps off the building to the impact of the water, it is roughly 3 seconds due to human reaction time. This equation neglects the drag that air causes, however.

Lastly, in order to calculate the final velocity of the man just before his impact with the water, plug the values into the following equation:

x = 2.6 + 9.8(2.57)

x = 2.6 + 25.18

x = 27.78

final velocity = 27.78 m/s

Now, we have a value for each comment of the man's jump into the harbor.

Initial Velocity: 2.6 m/s

Final Velocity: 27.78 m/s

Acceleration: 9.8 m/s^2

Time: ~2.57 seconds

Displacement: 39.3 meters

Baylor Wallace

October 21, 2016