# A bouncy ball is dropped from a height of 144 feet. The function h=−16x^2+144 gives the height, h, of the ball after x seconds. When does the ball hit the ground?

Feb 24, 2018

The ball will hit the ground in $3$ secs.

#### Explanation:

$16 {x}^{2}$ must equal $+ 144$ so that $h = 0$.

In other words, when the ball hits the ground, the distance above the ground will be zero.

That occurs $3$ seconds after the ball was released.

$0 = - 16 {x}^{2} + 144$

$16 {x}^{2} = 144$

Feb 24, 2018

$3$ seconds

#### Explanation:

$h \left(x\right) = - 16 {x}^{2} + 144$ => the ball hits the ground at h = 0:
$- 16 {x}^{2} + 144 = 0$
$- 16 \left({x}^{2} - 9\right) = 0$
${x}^{2} = 9$
$x = \pm 3$ => reject the negative time, thus:
$x = 3$ the ball hits the ground 3 seconds after it was dropped

Feb 24, 2018

The ball hits the ground after $3$ seconds.

#### Explanation:

When the ball is dropped, the height will get less and less. When it hits the ground the height will be $0$

$0 = - 16 {x}^{2} + 144 \text{ } \leftarrow$ now solve the equation,

You can isolate the $x$ term:

$0 = - {x}^{2} + 9 \text{ } \leftarrow \div 16$

${x}^{2} = 9$

$x = \pm \sqrt{9}$

$\textcolor{b l u e}{x = + 3} \mathmr{and} x = - 3$

You could also factorise:

$0 = 144 - 16 {x}^{2} \text{ } \leftarrow$ difference of two squares.

$0 = 9 - {x}^{2} \text{ } \leftarrow \div 16$

$0 = \left(3 + x\right) \left(3 - x\right)$

$x = - 3 \mathmr{and} \textcolor{b l u e}{x = + 3}$

In each case reject $- 3$ because the time cannot be negative,