Answer
a. $y=-16x^2+40x+200$.
b. $y=0$
The ball's height above the ground after $5$ second is $0$, meaning the ball is on the ground.
Work Step by Step
a.To write the quadratic equation $y=ax^2+bx+c$, first we need to get $a,b$, and $c$ by solving the mathematical model that describes the relationship for the ball of height above the ground.
The mathematical model is,
$x=1$, $y=224$ { $224=a+b+c$,
$x=3$, $y=176$ { $176=9a+3b+c$,
$x=4$, $y=104$ { $104=16a+4b+c$,
Let's take a pair of two equations and eliminate $c$,
{$224=a+b+c$ and {$176=9a+3b+c$
{$176=9a+3b+c$ , { $104=16a+4b+c$ ,
Multiply one of pair's equation by $-1$, to eliminate $c$
{$224=a+b+c$
$-1${{$176=9a+3b+c$, $->$ $-176=-9a-3b-c$
Adding the equation will result in the first pair of the equation being, $48=-8a-2b$.
doing the same to the second pair of the equation will result,
$-72=7a+b$
Arrange the result to eliminate the one variable and solve for one,
{$48=-8a-2b$
{$-72=7a+b$,
Multiplying the second equation by $2$ and adding the equation,
{$48=-8a-2b$
$2${$-72=7a+b$, $->$ $-144=14a+2b$,
$(48-144)=(-8=14)a+(-2+2)b$,
$-96=6a$,
$a=-16$,
back-substituting and solving for $b$,
$48=-8(-16)-2b$,
$48-128=-2b$,
$-80=-2b$,
$40=b$.
back-substituting and solving for $c$,
$224=-16+40+c$,
$224-24=c$,
$200=c$.
Therefore the equation will be,
$y=-16x^2+40x+200$.
b.
$y=-16(5)^2+40(5)+200$,
$y=-400+200+200$,
$y=0$
The ball's height above the ground after $5$ second is $0$, meaning the ball is on the ground.
