# A ball is thrown straight up from the top of a 80 foot tall building with an initial speed of 64 feet per second. The height of the ball as a function of time can be modeled by the function h(t)=-16t^2+64t+80. When will the ball be 144 feet above ground?

## When will the ball hit the ground?

Oct 26, 2017

after $2$ seconds

#### Explanation:

$h \left(t\right) = - 16 {t}^{2} + 64 t + 80$

factorise:

$h \left(t\right) = 16 \left(- {t}^{2} + 4 t + 5\right)$

in this example, $16 \left(- {t}^{2} + 4 t + 5\right) = 144$

$- {t}^{2} + 4 t + 5 = \frac{144}{16} = 9$

• solving for $t$ using the quadratic equation:

$- {t}^{2} + 4 t + 5 = 9$

$- {t}^{2} + 4 t - 4 = 0$

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

$= \frac{- 4 \pm \sqrt{16 - 16}}{- 2}$

$= \frac{- 4 \pm 0}{-} 2$

$= - \frac{4}{-} 2$ (repeated root)

$= 2$

$t = 2$

assuming the time is in seconds, it will take $2$ seconds for the ball to reach $144$ feet above ground.

• solving graphically:

red: $y = - {x}^{2} + 4 x + 5$
blue: $y = 9$

to find where $- {t}^{2} + 4 t + 5 = 9$, the point of intersection of both graphs is used - $\left(2 , 9\right)$

[$x$ in the function is $t$ time]

this means that after $2$ seconds, the ball will have reached $9 \cdot 16$ feet (see factorisation), or $144$ feet.