# A ball is tossed upward from the ground. Its height

A ball is tossed upward from the ground. Its height in feet above ground after t seconds is given by the function $h\left(t\right)=-16{t}^{2}+24t$. Find the maximum height of the ball and the number of seconds it took for the ball to reach the maximum height.
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Viktor Wiley
1) Concept:
The maximum height of the ball is the maximum value of h(t). The degree of the function $h\left(t\right)=-16{t}^{2}+24t$ is 2 and hence it is a quadratic function. The graphical representation of the quadratic function is a parabola. The coefficient of the ${t}^{2}$ 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 $a{x}^{2}+bx+c$. Find the values of a and b.
Step 2: Find the vertex using the formula $\left(\frac{-b}{2a},h\left(\frac{-b}{2a}\right)\right)$.
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 $h\left(t\right)=-16{t}^{2}+24t$ with $a{x}^{2}+bx+c$, we have $a=-16.b=24.$
Now, to find the x-coordinate of the vertex, substitute the values of a and b in $\frac{-b}{2a}$.
$\frac{-b}{2a}=\frac{-24}{2\left(-16\right)}$
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.
$\frac{-24}{2\left(-16\right)}=\frac{2×2×2×3}{2×2×2×2×2}=\frac{3}{4}$
Now, find $h\left(\frac{3}{4}\right)$. Substitute $\frac{3}{4}$ for t in the function $h\left(t\right)=-16{t}^{2}+24t$.
$h\left(\frac{3}{4}\right)=-16{\left(\frac{3}{4}\right)}^{2}+24\left(\frac{3}{4}\right)$
Again, write the numbers as product of their factors.
$h\left(\frac{3}{4}\right)=-2×2×2×2×\left(\frac{3×3}{4×4}\right)+2×2×2×3\left(\frac{3}{4}\right)$
Cancelling out the common factors in the numerator and the denominator, we have,
$h\left(\frac{3}{2}\right)=-2×2×2×2\left(\frac{3×3}{2×2×2×2}\right)+2×2×2×3\left(\frac{3}{2×2}\right)$
Evaluate using order of operations.
$h\left(\frac{3}{2}\right)=-3×3+2×3×3$
$h\left(\frac{3}{2}\right)=-9+18=9$
Now, we have,
$\frac{-b}{2a}=\frac{3}{4}$ and $h\left(\frac{-b}{2a}\right)=9$.
Therefore, the vertex is $\left(\frac{3}{4},9\right)$
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 $\frac{3}{4}$ or 0.75 seconds for the ball to reach the ground.