Question

99 POINT QUESTION

Answer 1

We assume you intend h(t) to be the usual equation of ballistic motion
  h(t) = -16t² + 40t + 10

a. The function can be written in vertex form as
  h(t) = -16(t² -5/2t) +10
  h(t) = -16(t² -5/2t +(5/4)²) + 10 +5²
  h(t) = -16(t -5/4)² + 35
This is the vertex form a(x-h)²+k where (h, k) is the vertex of the parabola. The maximum height will be "k" for a < 0, which it is.

The ball's maximum height is 35 ft.


b. h(t) = 0 for the value of t that satisfies
  0 = -16(t -5/4)² + 35
  (t -5/4)² = 35/16
  t = (5 +√35)/4 ≈ 2.729
About 2.729 seconds have passed when the ball hits the ground.


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I find it convenient to let a graphing calculator tell you the answers to these questions. As above, the maximum height is the y-coordinate of the vertex. The elapsed time is the x-coordinate of the positive x-intercept.

Answer 2

First of all, you left off your squared on the first x term.  The way you find the maximum height is to complete the square on the quadratic.  That vertex will give you your max height in the y value, and will give you the number of seconds it occurs at.  Begin by setting the quadratic equal to 0 and then moving over the constant: $-16 t^{2} +40t=-10$.  One of the rules for completing the square is that the leading coefficient has to be a positive 1.  Ours is a -16, so we have to factor it out: $-16( t^{2} -2.5t)=-10$.  Now we take half the linear term, square it, and add it to both sides.  Our linear term is 2.5.  2.5 divided in half is 1.25, and 1.25 squared is 25/16.  We will add it into the parenthesis on the left just fine, but don't forget that we have that -16 hanging out front still, so we are actually "adding" in -16*25/16, which will give us this: $-16( t^{2} - \frac{5}{2}t+ \frac{25}{16}) =-10-25$.  In this process we have created a perfect square binomial on the left.  That, along with doing the math on the right gives us $-16( t- \frac{5}{4}) ^{2} =-35$.  When we move the -35 back over by addition, we will get $-16(t- \frac{5}{4}) ^{2}+35=y$.  The vertex of this function is $( \frac{5}{4},35)$.  Therefore, at 1.25 seconds, the projectile reached its max height of 35 feet.  Now in order to find when the ball hits the ground, we will factor it to find the zeros.  Using the quadratic formula you will find that the zeros of the function are x = -.229 and x = 2.7.  The two things in math that will NEVER be negative are time and distance/length, so we know that we cannot have the time be the negative of those solutions.  Therefore, the projectile hits the ground at 2.7 seconds.