A ball is tossed upward from the ground. Its height

1) Concept:
The maximum height of the ball is the maximum value of h(t). The degree of the function ${\displaystyle h\left(t\right)=-16t^2+24t}$ is 2 and hence it is a quadratic function. The graphical representation of the quadratic function is a parabola. The coefficient of the ${\displaystyle t^2}$ is -16. Therefore, it is a parabola that opens downward. So, the h(t) or the y coordinate of the vertex gives the maximum height and the x coordinate of the vertex gives the number of seconds it took for the ball to reach the maximum height.
Step 1: Compare h(t) with the general quadratic equation ${\displaystyle a{x}^2+bx+c}$. Find the values of a and b.
Step 2: Find the vertex using the formula ${\displaystyle (\frac{-b}{2a},h\left(\frac{-b}{2a}\right))}$.
Step 3: The x coordinate of the vertex is the numbers of seconds the ball takes to reach the maximum height and the y coordinate of the vertex is the maximum height of the ball.
2) Calculation:
Comparing ${\displaystyle h\left(t\right)=-16t^2+24t}$ with ${\displaystyle a{x}^2+bx+c}$, we have ${\displaystyle a=-16.b=24.}$
Now, to find the x-coordinate of the vertex, substitute the values of a and b in ${\displaystyle \frac{-b}{2a}}$.
${\displaystyle \frac{-b}{2a}=\frac{-24}{2(-16)}}$
24 and 16 are multiples of 2. So,we can write them as factors of 2 and the common factors can be cancelled out. Since both the numerator and the denominator has negative sign, the fraction becomes positive.
${\displaystyle \frac{-24}{2(-16)}=\frac{2\times 2\times 2\times 3}{2\times 2\times 2\times 2\times 2}=\frac{3}{4}}$
Now, find ${\displaystyle h\left(\frac{3}{4}\right)}$. Substitute ${\displaystyle \frac{3}{4}}$ for t in the function ${\displaystyle h\left(t\right)=-16t^2+24t}$.
${\displaystyle h\left(\frac{3}{4}\right)=-16{\left(\frac{3}{4}\right)}^2+24\left(\frac{3}{4}\right)}$
Again, write the numbers as product of their factors.
${\displaystyle h\left(\frac{3}{4}\right)=-2\times 2\times 2\times 2\times \left(\frac{3\times 3}{4\times 4}\right)+2\times 2\times 2\times 3\left(\frac{3}{4}\right)}$
Cancelling out the common factors in the numerator and the denominator, we have,
${\displaystyle h\left(\frac{3}{2}\right)=-2\times 2\times 2\times 2\left(\frac{3\times 3}{2\times 2\times 2\times 2}\right)+2\times 2\times 2\times 3\left(\frac{3}{2\times 2}\right)}$
Evaluate using order of operations.
${\displaystyle h\left(\frac{3}{2}\right)=-3\times 3+2\times 3\times 3}$
${\displaystyle h\left(\frac{3}{2}\right)=-9+18=9}$
Now, we have,
${\displaystyle \frac{-b}{2a}=\frac{3}{4}}$ and ${\displaystyle h\left(\frac{-b}{2a}\right)=9}$.
Therefore, the vertex is ${\displaystyle (\frac{3}{4},9)}$
So, it took 0.75 seconds for the ball to reach the ground.
Conclusion:
The maximum height of the ball is 9 feet and it took ${\displaystyle \frac{3}{4}}$ or 0.75 seconds for the ball to reach the ground.