An acorn falls from the branch of a tree to the ground 25 feet below. the distance, s, that the acorn is from the ground as it falls is represented by the equation s(t) = –16t2 + 25, where t is the number of seconds. for which interval of time is the acorn moving through the air?

The end of its travel is when S(t)=0 or when it has hit the ground...

-16t^2+25=0  divide equation by -1

16t^2-25, this is a difference of squares which always factors:

(a^2+b^2)=(a+b)(a-b), in this case:

(4t+5)(4t-5)=0, since t>0

t=5/4=1.25, and since it started falling at t=0, its time moving in the air is:

1.25-0=1.25 seconds.

I’m pretty sure the slope would be 1 since the x and y values increase by 1