Stuntman Dan Koko jumped from the top of the Vegas World Hotel and Casino, a distance 99.4 meters above the ground, into an airbag.
How fast was he going when he hit the airbag?
Answer: There are two ways to find his final speed. Quick method: use the kinematic equation 2 2 v - vo = 2 * a * (x - xo) We know initial velocity vo = 0 displacement (x - xo) = -99.4 m acceleration a = -9.8 m/s^2 Solve for final velocity: 2 v = 2 * (-9.8 m/s^2) * (-99.4 m) v = sqrt ( 1948 m ) = 44 m/s The other method takes a little longer. First, we figure out how long it took him to fall to the airbag; then, we calculate how fast he was going when he reached it. Time to fall: use equation 2 (x - xo) = vo * t + 0.5 * a * t 2 -99.4 m = 0 + 0.5 * (-9.8 m/s^2) * t so t = sqrt ( 20.3 s^2 ) = 4.5 seconds Speed after falling for 4.5 seconds is v = vo + a * t = 0 + (-9.8 m/s^2) * (4.5 s) = 44 m/s One finds the same speed either way.
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