The university skydiving club has asked you to plan a stunt for an air show. In this stunt, two skydivers will step out of opposite sides of a stationary hot air balloon 2,500m above the ground. the second skydiver will leave the ballon 20 seconds after the first skydiver but you want them both to land on the ground at the same time. the show is planned for a day with no wind so assume all motion is vertical. to ger rough idea of the situation, assume that a skydiver will fall with a constant acceleration of 9.8m/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 3s after stepping out of the balloon before opening her parachute, How long must the second skydiver wait after leaving the balloon before opening his parachute?
I tried solving for time with the quadratic formula, but I am unsure of what procedure to take. Thanks for the help
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