Question

The height, h, in feet of a ball above the ground is given by h = −16t2 + 64t + 80, where t is the time in seconds. How long does it take the ball to hit the ground?

Answer 1

Answer:

t = 5 seconds

Step-by-step explanation:

It is given that, the height h, in feet of a ball above the ground is given by :

$h=-16t^2+64t+80$

Where t is in seconds

We need to find how long does it take the ball to hit the ground? When it hits the ground, its height will be equal to 0. So,

$-16t^2+64t+80=0\\\\t=\dfrac{-64\pm \sqrt{(64)^2-4\times (-16)(80)} }{2\times -16}\\\\t=\dfrac{-64+ \sqrt{(64)^2-4\times (-16)(80)} }{2\times -16},\dfrac{-64-\sqrt{(64)^2-4\times (-16)(80)} }{2\times -16}\\\\t=-1\ s\ \text{and}\ t=5\ s$

Neglecting negative time.

It means ball will take 5 seconds to hit the ground.