Question

How long does it take for the ball to hit the ground at -5t^2 + 10t + 320

Answer 1

if t is time, then we are solving for the value of t that makes the expression equals 0 (height of the ground)

0=-5t^2+10t+320

factor out -5

0=-5(t^2-2t-64)

use quadratic formula

for at^2+bt+c=0

`t=(-b \pm √(b^2-4ac))/(2a)`

in our case, a=1, b=-2, c=-64

`t=(-(-2) \pm √((-2)^2-4(1)(-64)))/(2(1))`

`t=(2 \pm √(260))/(2)`

`t=(2 \pm 2√(65))/(2)`

`t=1 \pm √(65)`

since
`1-√(65)<0` and we can't have negative time,

`t=1+√(65)` which is about 9.062 seconds