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?

Question

How to solve for time?

Answer 1

The second parachutist must open the parachute at the moment that she catches up with the first, i.e. has fallen the same distance as the first.

1. Establish the time/distance relation of first parachutist.
For the first 3 seconds,
distance = 0 + 9.8*(3²) = 44.1 m
total distance at time t (t≥3 s)
=44.1+3.2(t-3)
=3.2t+34.5

2. Establish the time/distance relation of second parachutist before her parachute was open (for t≥20):
distance=(1/2)9.8((t-20)²)
=4.9t²-196t+1960

Equate times and solve for time the two parachutists were at the same elevation
(note that time is measured from the first jump)
32t+345=49t²-1960t+19600
and solve for t.
Do subtract 20 seconds from t since the question asks for time since leaving the balloon. Reject all t<20 seconds.