Question

A model jet is fired up in the air from a 18-foot platform with an initial upward velocity of 56 feet per second. The height of the jet above ground after t seconds is given by the equation h=-16t^2+56t+18, where h is the height of the jet in feet and t is the time in seconds since it is launched. What is the maximum height the jet reaches, to the nearest foot? A. 767 feet B. 67 feet C. 30 feet D. 18 feet

Answer 1

The way you find the max height of this journey is by completing the square.  When you complete the square, you find the vertex of the parabola.  The vertex will give the max height (k), and the time at which it reached its max height (h).  The coordinates for the vertex are (h, k).  Completing the square requires that we set the function equal to zero and then move the constant over to the other side of the equals sign.  That will give us this: $-16 t^{2} +56t=-18$.  Now we can complete the square on the x terms.  The first rule is that the leading coefficient be a positive 1.  Ours is a -16, so we have to factor it out. $-16( t^{2} - \frac{7}{2} t)=-18$.  Now we will take half the linear term, square it, and add it to both sides.  Our linear term is 7/2.  Half of that is 7/4, and squaring 7/4 gives us 49/16.  We will add that in to the parenthesis on the left side just fine, but don't forget about the -16 hanging around out front.  What we have actually "added" in is -16*49/16, which gives us -49.  This is what we have now: $-16( t^{2} - \frac{7}{2}t+ \frac{49}{16})=-18-49$.  In that process, we have created a perfect square binomial.  That, along with doing the math on the right gives us $-16(t- \frac{7}{4}) ^{2}=-67$.  We will now move that -67 over by addition, and then we will have the vertex.  $-16(t- \frac{7}{4}) ^{2} +67=y$  The vertex of this parabola is (7/4, 67).  This means that at 11 3/4 seconds, the object reached its max height of 67 feet.  Choice B from above.