# A ball is thrown straight up in the air at a velocity of 24 ft./sec. The height of the ball at x seconds is given by the formula: h = 24x – 16x^2. How do you find the maximum height of the ball and when it will land?

May 7, 2018

Solve the quadratic formula to find the time for the ball to land, and then use half of that time to calculate the maximum height. You will find that the time to fall is 1.5 seconds and the maximum height is 9 feet.

#### Explanation:

First, lets solve the quadratic equation to determine the times when $h = 0$, or when the ball is on the ground. This will tell us the time that the ball will land, assuming the initial throw is considered as $t = 0$.

First, rewrite the function as a quadratic that has $h = 0$:

$0 = - 16 {x}^{2} + 24 x + 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = - 16$
$b = 24$
$c = 0$

$x = \frac{- 24 \pm \sqrt{{24}^{2} \cancel{- 4 \left(- 16\right) \left(0\right)}}}{2 \left(- 16\right)}$

$x = \frac{- 24 \pm \sqrt{{24}^{2}}}{- 32}$

$x = \frac{- 24 \pm 24}{- 32} \Rightarrow x = \left\{\frac{- 24 + 24}{-} 32 , \frac{- 24 - 24}{-} 32\right\}$

$\textcolor{g r e e n}{x = \left\{0 , 1.5\right\}}$

We already know that $x = 0$ is our starting point, so the time to land is the second root, 1.5 seconds.

As for the time to maximum height: We can assume that the ball hits the maximum height halfway through its travel since the ball is released from $h = 0$ and ends at $h = 0$.

If the time to maximum height is $\frac{1.5}{2} = 0.75 = \frac{3}{4}$ seconds, the height can be calculated:

$h = 24 x - 16 {x}^{2}$

$h = 24 \left(\frac{3}{4}\right) - 16 {\left(\frac{3}{4}\right)}^{2}$

$h = 24 \left(\frac{3}{4}\right) - \cancel{16} \left(\frac{9}{\cancel{16}}\right)$

$h = 6 \times 3 - 9$

$h = 18 - 9 = \textcolor{g r e e n}{9 \text{ feet}}$