Two skydivers will step out of opposite sides of a stationary hot air balloon 50

Question

Two skydivers will step out of opposite sides of a stationary hot air balloon 5000 feet above the ground. the second skydiver will leave the balloon 20 seconds after the first skydiver, but they will land on the ground at the same time. The skydivers fall at a constant acceleration of 32 feet/sec^2 before the parachute opens. As soon as the parachute opens the skydiver will fall at a constant velocity of 10 feet/sec. the first skydiver waits 3 seconds after leaving the balloon to open their parachute, how long must the other skydiver wait before opening his?

Explanation / Answer

First, you need to write an equation for the position of the first skydiver as a function of time.

This isn't difficult. It's a quadratic for the first 3 seconds, but after t=3 it's linear. I'd write it like this:

H1(t) = 1500 - ½ 9.80 t², 0 <= t < 3
H1(t) = H1(3) - 3.2 (t - 3.0), t >= 3

The position of the second skydiver starts at 1500 m at t=20 and accelerates downward until the 'chute opens:

H2(t) = 1500 - ½ 9.80 (t-20)², t=20 until chute opening

If the two are to land together, 'chutist 2 should open her 'chute when H2 = H1. Set H2(t) = H1(t) in the period t > 20 and solve for t. The opening delay is the difference between that time and 20 seconds.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.