Abaseball is thrown straight upward with an initial speed of 64 ft/sec. the number of feet s above the ground after t seconds is given by the equation x16+641. at what times will the baseball be 48 ft above the ground?

We are given

A baseball is thrown straight upward with an initial speed of 64 ft/sec

and

The number of feet s above the ground after t seconds is given by the equation -16t^2 +64t

so, we get

we can set s(t)=48

and then we can solve for t

we can factor out 16

divide both sides by 16

and we get

we can multiply both sides by -1

now, we can factor it

So, the baseball be 48 ft above the ground at 1 sec and 3 sec.......Answer