Answer
a. $1$ second and $1.5$ seconds
b. the ball never reaches $48$ ft
c. $h=25$ ft
d. $1.25$ seconds
e. $2.5$ seconds
Work Step by Step
$h=-16t^2+v_0t$
a.
$v_0=40\frac{ft}{s}$
$24=-16t^2+40t$,
$0=-16t^2+40t-24$,
$0=-2t^2+5t-3$, dividing both sides by $8$.
$0=-2t^2+2t+3t-3$,
$0=-2t(t-1)+3(t-1)$,
$0=(-2t+3)(t-1)$,
thus, $-2t+3=0, -2t=-3, t=\frac{3}{2}$ or $t-1=0, t=1$.
This means that the ball reaches $24$ ft at $x=1$ when the ball is going upward to reach maximum height and at $x=1.5$ when the ball is going down after passing maximum height.
b.
$48=-16t^2+40t$,
$16t^2-40t+48=0$,
$2t^2-5t+6=0$,
See for the determinant before solving for the trinomial to see if the trinomial has a solution or not.
if $b^2-4ac\gt0$, the trinomial has many solutions,
if $b^2-4ac=0$, the trinomial has one solution,
if $b^2-4ac\lt0$, the trinomial has no solution.
Thus, In this case, $(-5)^2-4(2)(6)=-23$, therefore the trinomial has no solution. Meaning, In this case, The ball never reaches the height of $48$ ft.
c.
By turning the polynomial from a standard form into a vertex form,
$h=-16t^2+40t$, multiply both sides by $-1$
$-h=16t^2-40t$, add $25$ to both sides,
$-h+25=16t^2-40t+25$,
$-h+25=(4t-5)^2$, substract $25$ from both sides,
$-h=(4t-5)^2-25$, multiply both sides by $-1$,
$h=-(4t-5)^2+25$
Therefore, we can see from the trinomial that $h=25$ is the maximum height.
d.
$h=-(4t-5)^2+25$ is $h=25$ at $t=\frac{5}{4}$
e. $h=-16t^2+40t=t(-16t+40)=0$,
either $t=0$ or $-16t+40=0, t=2.5$.
Thus, $h=0$ at $t=0$ when the ball is first thrown and $h=0$ at $t=2.5$ when the ball hits the ground.
Thus, the ball hits the ground at $t=2.5$.
