Answer
a.maximum height is $33.7$ and it occurs at $x$ of $26.25$
b.maximum horizontal distance is $55.3$
c.from the height of $6.1$ was the shot released
Work Step by Step
$g(x)=-0.04x^2+2.1x+6.1$
By completing the square and turning an equation from the standard form into a vertex form can we get the maximum and the horizontal distance it occurs at.
To complete the square, we need to turn an equation into a perfect square
$g(x)=-0.04(x^2-52.5x+(-52.5/2)^2)+6.1=-0.04(x-52.5/2)^2+6.1$,
To balance an equation to equal the previous one, as we added $(-0.04\times(-52.5/2)^2)$ into the equation, we need to substract $(-0.04\times(-52.5/2)^2)$ from the equation.
$-0.04(x-52.5/2)^2+6.1-(-0.04\times(-52.5/2)^2)=-0.04(x-52.5/2)^2+6.1-(-27.5625)=-0.04(x-52.5/2)^2+33.6625$
a.from the equation we can see that the maximum height of the ball is $33.6625$ and it happens at $x=52.5/2=26.25$
b.the maximum horizontal distance of the ball is,
$-0.04x^2+2.1x+6.1=0$
using the quadratic formula, $\frac{-b+or-\sqrt {b^2-4ac}}{2a}$. we can get an answer of $x=-2.76 or x=55.26$. since we're finding solution for a distance here, only the positive answer is relevant. therefore $x=55.3$
c.$g(x)=-0.04x^2+2.1x+6.1$
the initial height of the ball occurs at $x=0$, $g(0)=6.1$. therefore the intial height of the ball is $6.1$
