Question

A baseball 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?

Answer 1

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

$s(t)=-16t^2+64t$

we can set s(t)=48

and then we can solve for t

$48=-16t^2+64t$

we can factor out 16

$48=16(-t^2+4t)$

divide both sides by 16

and we get

$3=(-t^2+4t)$

$-t^2+4t-3=0$

we can multiply both sides by -1

$t^2-4t+3=0$

now, we can factor it

$(t-1)(t-3)=0$

$(t-1)=0$

$t=1$

$(t-3)=0$

$t=3$

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