This one is a little tricky, but let's walk through it. We are trying to find 'x', the distance above the ground when The Rocketeer should catch up and start decelerating with the rocket pack. Then we can call the distance that the student falls freely (553 - x) because the two distances obviously add up to the height of the tower. The key equation that we are going to use is:
(v_f)^2 = (v_i)^2 + 2*a*d
where v_f is final velocity, v_i is initial velocity, a is acceleration, and d is distance. If you are not familiar with this equation, notice that you can get it if you take the definition of acceleration: v_f = v_i + a*t and square both sides, and then plug in d from the equation d = v_i*t + (1/2)*a*t^2
Anyway, for the first part of the fall we can plug in to the above equation to get (v_f)^2 = 2*g*(553-x) since v_i is zero.
The second half of the fall (with The Rocketeer) is described by this equation:
0 = (v_i)^2 + 2*(-5*g)*x or (v_i)^2 = 10*g*x
where the acceleration is negative because it is in the other direction.
Now comes the key, notice that v_f in the first equation is the same thing as v_i in the second equation. Whatever speed the student gets to at the end of the free fall is the speed that he will be starting with when The Rocketeer arrives. So we can set our two equations equal to each other:
10*g*x = 2*g*(553-x)
The 'g's cancel out, and we can distribute the 2 across:
10*x = 1106 - 2*x
Solve for x:
x = 92.2 meters
Now that we've gone through all that math, it's worth noting a "tricky" way to do the problem. Just realize that however far the student falls freely under gravity, he will only have to fall one-fifth as far to land perfectly on the ground, because The Rocketeer makes them slow down at exactly five times the speed of gravity. So if the student falls five sixths of the way down the building and The Rocketeer helps out for the last sixth they will land perfectly. And one sixth of the building height is in fact exactly what we just solved for.