Analysis+Calculations

 This analysis page includes the following: -Experimental height of the airbag at the top of its path -Actual height of the airbag at the top of its path -Time of flight -Velocity and Acceleration of the airbag -Potential and Kinetic energy -Impulse and momentum of the airbag -Force of the airbag

 ANALYSIS CALCULATIONS USING DATA:

Velocity: 18.83 m/s (from slope of graph) --> (1ft/.3048 m) (1mi/5280ft) (60s/1min) (60min/1hr) = 42.12 mi/hr Final Velocity: 0m/s Acceleration : -9.8m/s^2 v^2 = Vo^2 + 2ax 0 = (18.83^2) + 2(-9.8)x -354.57 = -19.6x x (calculated height) = 18.09m
 * Finding the maximum height with initial velocity from slope of graph

Actual Height = 9.07 m v^2 = vo^2 + 2ax 0 = vo^2 + 2(-9.8)(9.07) vo^2 = 177.772 vo = 13.33 m/s --> (1ft/ .3048 m) (1mi/ 5280ft) (60 s/ 1min) (60 min/ 1 hr) = 29.82 mi/hr


 * Our calculated height was inaccurate because our initial velocity being wrong. This was probably because the initial velocity depended on the graph on logger pro, and the scaling was off in the video because the meter stick, being so far away, was difficult to see.

Initial Velocity of graph - 18.83 m/s Calculated Initial Velocity - 13.33 m/s (18.83 - 13.33) / ((18.83 + 13.33)/2) = .342 * 100 = 34.2% v = vo + at 0 = 18.86 + -9.8t -18.86 = -9.8t t = 1.92s (until it reaches the top of its arc) t = 3.85s until the airbag hits the ground again.
 * Percent Difference between initial velocities
 * Finding the time

Mass of airbag: 2 kg mgh = (2)(9.8)(9.07) PE = 177.8J at the top.
 * Finding potential energy at the top

1/2mv^2 = 1/2 (2) (13.33)^2 KE = 177.7J initially
 * Finding kinetic energy

Percent difference between kinetic and potential energy. Potential energy: 177.8 J Kinetic energy: 177.7 J (177.8 - 177.7) / ((177.8 + 177.7)/2) = 0.1 / 77.75 = .000005% difference

Fexplosion - Fgravity = ma Fg = (2kg)(71.72)^2 Fg = 143.44N Fe = ma + mg Fe = (2kg)(71.72)^2 + (2kg)(-9.8m/s^2) Fe =123.84N
 * Finding the force

p=mv p=(2kg)(18.83m/s) p= 37.66Ns
 * Finding the Momentum of the Airbag

J = F∆t J = (123.84) * (0.2) = 24.768 Ns J = mv -mv J = (2)*(18.83) = 37.66 Ns
 * Finding the Impulse

Percent difference between force*time and change in momentum. Impulse: 24.768 Ns Change in time: 37.66 Ns (37.66 - 24.768) / ((37.66 + 24.768) / 2) = .413 * 100 = 41.3% difference 