Good,+Bad,+and+Ugly+Physics

**__Good__** The movie //Titanic// shows quite a few people jumping and falling off of the sinking ship. There was one instance where someone fell off of the tip of the //Titanic// and landed in the water below.

Average amount of time for the person to hit the water that I initially recorded: 4.4s

t: ? Vi: 0 m/s a= -9.8 m/s 2 d: 93.18 m //* The Titanic// measured 269 m. In this particular scene, only two fifths of the entire length of the boat was above water (the other three fifths had already been submerged.) The ship was approximately at an angle of 30° above sea level. This means that the vertical distance from the water to the tip of the ship is 93.81 m.



In order to find the time it should realistically take the person to hit the water, I used the equation d=Vit+½at2. d=½at2 2d= at2 t=√(2d/a) t=√(2*-93.18/-9.8) =4.36s The movie was accurate about the affects of gravity on people falling from the ship. <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">[|Video Clip]

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">**__Bad__** <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">In the movie //Taxi//, Queen Latifa tricks Gisele into jumping her car over a huge gap and trapping her on a small stretch of a bridge. I will assume that Gisele was traveling at about 90 mph because this is a pretty intense scene. 90 mph converts to 40.23 m/s. The distance that the car jumped was 11.2 times the car length. The average length of a four-door compact car is 4.12 m. This would mean that the jump was approximately 46.12 m. This jump also had virtually no incline, but I decided to give them a 5° incline for the benefit of the doubt.


 * . || x || y ||
 * V i || 40.08 m/s || 3.51 m/s ||
 * V f || 40.08 m/s || -- ||
 * a || 0 m/s 2 || -9.8 m/s 2  ||
 * d || 46.12 m || -2.44 m ||
 * t || 1.15 s || 1.15 s ||

V ix =V i cos ө40.23cos5 40.08 m/s V iy =Visin ө 40.23sin5 3.51 m/s

D x = __(V i + V f ) t__ 2 2 Dx = (V i + V f ) t

__2 D x __ = t (V i + V f )

t = __2 (46.12)__ __.__ = 1.15 (40.18+40.08)

d = Vi t + ½at 2 d = 0 + ½(-9.8)(1.15) 2 = -2.44

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">This means that by the time the car should be reaching the other side of the jump, it will have already dropped 2.44 m below the landing. [|Video Clip] (2:03-2:10)

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">**__Ugly__** <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">In the final ice dance of the movie //Blades of Glory//, Will Farrell and Jon Heder go up for a manly chest bump and fly backwards continuing their routine. I did a little research and found out that Will Farrell weighs approximately 200 lbs, which converts into 90.72 kg, and Jon Heder weighs approximately 175 lbs, which converts into 79.38 kg. I also found that the average ice skater will go into a jump at about 30 mph, which is 13.4 m/s. Looking at this clip, I was able to assume that the height that these two men made contact was about 1.25 m off of the ice.

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">First, I needed to find the momentum of each person <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">P<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">WillInitial <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">﻿= mV = (90.72)(30) = 2721.6 <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">PJ<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">onInitial = mV = (79.38)(-30) = -2381.4

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">Next, I needed to find the final momentum of each person <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">P<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">WillFinal = P<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">WillInitial + P<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">JonInitial = 2721.6 + -2381.4 340.2 <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">P<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">JonFinal = P<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">JonInitial + P<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">WillInitial = -2381.4 + 2721.6 340.2

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">Then, I found the velocity of each person after contact. <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt 0.5in;">P = mV ; V = P/m <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">V<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">Will =P/m = 340.2/90.72 = 3.75 m/s <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">V<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 80%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: sub;">Jon =P/m = 340.2/79.38 = 4.3 m/s

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">Next, I found the amount of time that they were supposed to be in the air for. <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">dy = 1.25 m <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">Viy = 0 m/s <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">a = -9.8 m/s 2 <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 90%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: super;">t = ? <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">d = Vi t + ½at<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 70%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: super;">2 <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">d = ½at<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 70%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: super;">2 <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">2d = at<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 70%; margin: 0in 0in 0pt; text-indent: 0.5in; vertical-align: super;">2 <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">t = √(2d/a) <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">t = √(2(1.25)/-9.8) = .51 s <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt 1in;">**The movie is already inaccurate because they were in the air for over one full second.

<span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">Lastly, I found the total distance in the x-direction that Jon should have covered. (We do not see Will land in the movie.) <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt; text-indent: 0.5in;">V = d/t ; d = Vt = (4.3)(.51) = 2.19 m opposite of the direction he started. <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">In the movie, Jon lands well over 2.19 m away from the point of contact. <span style="font-family: Tahoma,Geneva,sans-serif; margin: 0in 0in 0pt;">[|Video Clip] (0:22-0:28)