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Human Cannonball Launch: Safety, Thrill, and Engineering Precision

Experience the thrill of the human cannonball! Our engineering team has calculated a launch angle of 45 degrees and a safe velocity of 40 m/s, ensuring an exhilarating yet safe flight through a blazing hoop and into a safety net 2 meters above the ground. The human cannonball will achieve peak flight at 40 meters high, passing through the hoop at 2.9 seconds, 81 meters from the cannon. Watch the stunning spectacle unfold, combining the perfect blend of safety and excitement, designed to amaze the crowd while minimizing risk.

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Human Cannonball Launch: Safety, Thrill, and Engineering Precision

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  1. Human Cannonball Isaac Carluccio and Kelly Pitner

  2. Given: • Our engineering team has chosen a 45 degree angle of release because the initial vertical and horizontal velocities will be the same. • Our firing velocity is below the amount that would kill someone, but plenty high enough to have the biggest thrill. Therefore, we have chosen a velocity of 40 m/s. • In order to make sure the human cannonball is safe, the human will soar out of the cannon, through a blazing hoop correctly placed and land in the safety net two meters above the ground to ensure that there are minimal injuries throughout the entire flight.

  3. Calculations: • Time overall: yf = yi +Viy(t) + ½(-9.8)(t^2) 2 = 0 + 40sin(45)(t) - 4.9t^2 0 = -4.9(t^2) + 28(t) - 2 (Use Quadratic Formula) A = -4.9 B = 28 C = -2 T = 6 seconds

  4. Calculations: • Distance for Safety Net: Initial Velocity in the horizontal = 40cos(45) =28m/s Time=6s Vx= delta x/ delta t Vx* delta t = delta x 28 * 6 = delta x Delta x = 168 meters

  5. Calculations: • Time where human cannonball would pass through hoop or peak of trajectory: Vfy = Viy + (-9.8)t where Vfy is the peak of the trajectory therefore Vfy = 0 m/s (Vf – Vi)/a = time at peak (0-28)/-9.8 = time at peak Time at peak = 2.9 seconds

  6. Calculations: • Placement of hoop: You would use Vx = delta x / delta t Vx * delta t = delta x or placement of the hoop 28 * 2.9 = delta x Delta x = 81 meters • Height of hoop: Vfy^2 = Viy^2 + 2(a)(delta y) Vfy^2 – Viy^2 / 2(a) = delta y where Vfy = 0 m/s at peak -28^2 / 2(-9.8) = delta y Delta y = 40 meters

  7. Proposal: • When firing the human cannonball at 40 m/s at an angle of 45 degrees, we determined the human cannonball would pass through the blazing hot hoop at 2.9 seconds while 40 meters in the air at a position 81.2 meters from the cannon. Continuing to the safety net 3.1 seconds later approximately 86 meters further from where he passed through the hoop. To the crowd, this will look completely effortless since all the calculations and work is already done for you. For the most amazing, thrilling, and eye-popping event, we believe our product is the one for you!

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