Question

Draw position, velocity, acceleration graphs for Lin Lee and Rodrigo and also label an image of this scenario with velocity and acceleration vectors. The 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 basket held up by a stationary hot air balloon 5,000 feet above the ground. The second skydiver will leave the basket 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 you may assume all motion is vertical. To get a rough idea of the situation, assume that a skydiver will fall with a constant acceleration of 9.8 m/sec2 before the parachute opens. As soon as the parachute is opened, the skydiver falls with a constant speed of 3 m/sec. If the first skydiver, Lin Lee, waits 3 seconds after stepping out of the balloon before opening her parachute, how long must the second skydiver, Rodrigo, wait after leaving the balloon before opening his parachute?

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.
5000ft = 1524m
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(t-3)
=3t+35.1

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)
30t+351=49t²-1960t+19600
and solve for t.

t = 24.7s or15.89s
Do subtract 20 seconds from t since the question asks for time since leaving the balloon. Reject all t<20 seconds(hence neglecting 15.89s)

t= 24.7- 20 = 4.7s

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