Question

7. 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 hotair 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 land on the ground at the same time. The show is planned for a day with no wind so assume that all motion isvertical. To get a rough idea of the situation, assume that a skydiver will fall with a constant acceleration of 9.80 m/s2before the parachute opens. As soon as the parachute is opened, the skydiver falls with a constant velocity of 3.2 m/s. If the first skydiver waits 3 seconds 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?

Answer 1

Find an equation for the distance the first jumper falls. You can calculate how far the first falls before pulling open her parachute using:

D=0.5*g*t^2

D = 0.5*9.8*3^2=44.1 m

Then you have a constant velocity function with a slope of -3.2.

Position of first jumper:

P = 1500-44.1-3.2*t = -3.2*t+1455.9

That has the initial height of 1500 meters, minus the distance the jumper fell before pulling open her chute, then a constant fall speed.

Since the other person waits 20 seconds before jumping, add on another 17 seconds of fall time to the first person’s equation, this lets you make two equations with the same starting value for time:

P = -3.2*17+1455.9 = -3.2*t+1401.5

Since we want them to land at the same time, and they will both fall at the same rate when their chutes are both open, we want a function for the falling distance of the second jumper:

P = -0.5*g*t^2 + 1500

= -0.5*9.8*t^2 + 1500

= -4.9*t^2 + 1500

This is a formula for the second jumper in the same terms of time as the first, set the two equal to each other and solve for time t:

-4.9*t^2 + 1500 = -3.2*t+1401.5

0=-4.9*t^2+3.2*t+98.5

Use the quadratic formula and solve for t:

t = 4.82 seconds