It is my hope that, as soon as you’re done figuring this problem out, you’ll be on your way to a wonderful (and professionally arranged and totally safe) fireworks display. What follows is not an SAT question, but if you enjoy math enough to visit my blog regularly and try these challenge questions, my hope is that you’ll think it’s fun anyway.

The prize this week: you’ll be close enough to the fireworks to feel the explosions in your gut. I love that. If you don’t love that, your prize can be different. I’m easy!

The height, in feet, of a pyrotechnic rocket t seconds after the fuse is lit is given by the function above, where r is a constant. If the rocket explodes 6.5 seconds after the fuse is lit at a height of 324 feet, for how many seconds does the fuse burn before the rocket takes off?

Put your answers in the comments. I’ll post the solution on Tuesday. Happy 4th of July, if you’re in the States. 🙂

UPDATE: Solution below the cut.

I think the fastest way to get to a solution here is to recognize that every time t appears in the function, it’s accompanied by ” – r“. If you’re really kicking butt on function translation, you might even recognize that the constant r is applying a rightward shift to the parabolic function.

Quick and dirty: substitute a single variable (I’m going to use x) in for (t – r):

y = 144x – 16x2
And graph it. Use your fancy calculator to calculate the maximum, which is at x = 4.5, y = 324.
If t = 6.5 when the rocket hits 324 feet, then 4.5 must equal 6.5 – r, and therefore r = 2.
So the answer I’m looking for, the time it takes for the fuse to burn down, is 2 seconds. That’s the rightward shift of the graph, which causes the rocket not to take off right at t = 0. Take a look at the graph as an illustration of the position of the rocket above the ground as a function of time (remembering that in our piece-wise function, all of the values less than 2 would actually just be zero — the rocket is just sitting on the ground):
Of course, there’s also an algebraic solution: substitute 324 in for h(t), substitute 6.5 in for t, and solve for r. It’s a bit more labor intensive, but it’s a fine way to go, especially if you don’t use a graphing calculator. You still, of course, need to recognize that r is the fuse time.

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