Object shot upward. Distance in feet above ground after t secs is s(t)=-16t^2+64t+80, how many secs to ground?

Answer 1

I will give it a go. You want to solve

-16*t^2 + 64*t +80 = 0

We can factor this as follows.

-16(t^2-4*t-5)=-16*(t-5)(t+1)=0

So the roots are t=-1 (extraneous solution) and t = 5.

The answer is 5 seconds.

Answer 2

s(t) = - 16t² + 64t + 80
16t² - 64t - 80 = 0
t² - 4t - 5 = 0
(t - 5).(t + 1) = 0
t = 5
After 5 secs. object is at ground level.

Answer 3

Might be homework, but the quadratic formula should give a nice simple result.

(That is if the scratchpad in my brain is still working.)

Answer 4

again i feel like this may be homework