Question

Two Skydivers jump out of a plane The first skydiver's motion is g(t)=4000-5(t+10)² The second skydiver jumps 10s later The motion of the second skydiver is h(t)=4000-5r² For both functions the height is measured in metres and the time is the number of seconds after the second skydiver jumps a)graph the function on the same set of axes b)will the second skydiver catch up to the first skydiver before the have to open the parachute at 800m  (THIS IS THE FULL QUESTION)

Answer 1

There is an error in the question. h(t)=4000-5t^2 not 4000-5r^2.

Let's analyse the functions. When t=0 (for the 2nd skydiver) the 1st skydiver has already jumped 10 seconds earlier, as shown by t+10 in the function. At t=0 the second skydiver is at 4000m while the first has already fallen 500m because g(0)=4000-500=3500m.

When g(t)=800, 4000-5(t+10)^2=800 therefore (t+10)^2=640, t+10=±8√10 from which t=8√10-10=15.3 seconds approx. We ignore the negative t. This time is from the 2nd skydiver's point of view, and h(15.3)=2829.8m approx. So the 2nd skydiver is more than 2000m higher than the first when the first's parachute is deployed.

The graphs are each part parabolas, because we can only look at f(t) and g(t) when t≥-10. t<-10 has no meaning in this context. The parabolas are parts of an inverted U shape with vertices at 4000. The units on the vertical axis (height) are in 100's of metres. The parabola on the left in red is the first skydiver. The green horizontal line is 800m above the ground where the parachutes are deployed. Consider only the falling part of the parabolas. t=-10 is the time when the first skydiver jumps.