Question

A ball is thrown downward from the top of a 240​-foot building with an initial velocity of 12 feet per second. The height of the ball h in feet after t seconds is given by the equation h=-16t^2 - 12t + 240. How long after the ball is thrown will it strike the​ ground?

Answer 1

Answer:

$16t^2 +12 t -240 =0$

And we can use the quadratic formula given by:

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

Where $a = 16, b = 12, c =-240$. And replacing we got:

$t = \frac{-12 \pm \sqrt{(12^2) -4*(16)(-240)}}{2*16}$

And after solve we got:

$t_1 = 3.516 s , t_2 = -4.266s$

Since the time cannot be negative our final solution would be $t = 3.516 s$  

Step-by-step explanation:

We have the following function given:

$h(t) = -16t^2 -12 t +240$

And we want to find how long after the ball is thrown will it strike the​ ground, so we want to find the value of t that makes the equation of h equal to 0

$0 = -16t^2 -12 t +240$

And we can rewrite the expression like this:

$16t^2 +12 t -240 =0$

And we can use the quadratic formula given by:

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

Where $a = 16, b = 12, c =-240$. And replacing we got:

$t = \frac{-12 \pm \sqrt{(12^2) -4*(16)(-240)}}{2*16}$

And after solve we got:

$t_1 = 3.516 s , t_2 = -4.266s$

Since the time cannot be negative our final solution would be $t = 3.516 s$