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| Emily's Measurement Lab! |
| File Size: | 146 kb |
| File Type: | doc |
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Human Cannonball
Calculations
Initial Angle = 52.5 degrees Initial velocity = 20 m/s Net Placement: vx = v*cos(theta) vx = 12.175 m/s vy = v*sin(theta) vy = 15.867 m/s time = (change in x)/(v*cos(theta)) time = (change in x)/12.175 s a = -9.8 m/s^2 delta x = vx*t t = (delta x)/vx delta y = vy*t+1/2*a*t^2 delta y = 15.867m/s*(delta x)/(12.175m/s)-4.9*((delta x)^2)/(148.2382m^2/s^2) delta y = 1.3032*delta x - .03306*(delta x)^2 0 = 1.3032*delta x - .03306*(delta x)^2 delta x = 0m and delta x = 39.4252m The net should be 39.4252m away at the same initial hight from which the human cannonball is fired. Hoop placement: delta x = (39.4252m)/2 = 19.7126 m t = (delta x)/vx t = 19.7126m/(12.175m/s) t = 3.2381 s delta y = 15.867m/s*t-4.9*t^2 delta y = 15.867m/s*3.2381s-4.9m/s^2*(3.2381s)^2 delta y = 12.8451 m The center of the hoop should be 19.7126 meters away in the x direction and 12.8451 meters away in the y direction.
