2 skydivers step out of opposite sides of a stationary hot air balloon 1500 mete

Question

2 skydivers step out of opposite sides of a stationary hot air balloon 1500 meters above the ground. The second skydiver will leave the balloon 20 seconds after the first skydiver but you want them both to hit the ground at the same time. With no wind and assume all motion is vertical. To get a rough idea of the situation, assume that a skydiver will fall with the constant acceleration of 9.80m/s^2 before the parachute opens. As soon as the parachute is opened, the skydiver falls with a constant velocity of 3.2m/s. If the first skydiver waits 3 seconds after stepping out of the ballon before opening her parachute, how long must the second skydiver wait after leaving the ballon before opening his parachute? Please show as much work as possible.

Explanation / Answer

1500 = 1/2 g t1^2 + 3.2 t2 1500 = 1/2 g * 3^2 + 3.2 t2 for the first diver t2 = 455 sec so total time T = 458 sec T for second diver = 458 - 20 = 438 sec 1500 = 1/2 g t^2 + 3.2 * (438 - t) where t is the time the second diver is in free fall Solve the quadratic for t

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