Answer:
a) 51 feet
b) 3.5 seconds
Step-by-step explanation:
The y-coordinate of the vertex of the given parabola is what we are looking for.
We first need to find the t-coordinate of the vertex.
The t-coordinate can be found using -b/(2a).
We need to compare
-16t^2+56t+2
to
at^2+bt+c
to identify a,b, and c.
a=-16
b=56
c=2
We are ready to compute -b/(2a).
-b/(2a)=-56/(2*-16)=-56/-32=7/4.
The vertex occurs at t=7/4.
To find y, we use y=2+56t-16t^2
y=2+56(7/4)-16(7/4)^2
y=51
So the maximum height is 51 feet.
Part b)
Hitting ground means the height between the ball and the ground is 0.
So we need to replace h(t) with 0.
0=2+56t-16t^2
I'm going to use quadratic formula.
a=-16
b=56
c=2
The quadratic formula is:
[tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{-56 \pm \sqrt{56^2-4(-16)(2)}}{2(-16)}[/tex]
Computing the thing inside the square root and the thing in the denominator using my handy dandy calculator:
[tex]t=\frac{-56 \pm \sqrt{3264}}{-32}{/tex]
I'm going to do the square root of 3264 now:
[tex]t=\frac{-56 \pm 57.131427428}{-32}[/tex].
I'm going to compute both
[tex]\frac{-56 + 57.131427428}{-32}[/tex] and [tex]-56-57.131427428}{-32}[/tex] using my handy dandy calculator:
[tex]-0.035357[/tex] while the other one is [tex]3.535357[/tex]
The negative value doesn't make sense for our problem so the answer is approximately 3.5 seconds.
