we are given
[tex]s(t)=-4.9t^2+20t+60[/tex]
(a)
we can plug t=0
[tex]s(0)=-4.9(0)^2+20(0)+60[/tex]
[tex]s(0)=60[/tex]
so, the height above the ground when the object is launched is 60 meter......Answer
(b)
we can set s(t)=0
and then we can solve for t
[tex]s(t)=-4.9t^2+20t+60=0[/tex]
[tex]-4.9t^2+20t+60=0[/tex]
we can use quadratic formula
[tex]t=\frac{-200-\sqrt{200^2-4\left(-49\right)600}}{2\left(-49\right)}[/tex]
we get
[tex]t=6.0917seconds[/tex]..........Answer
(c)
we can find vertex
[tex]t=\frac{-20}{2*-4.9}[/tex]
t=2.040816
now, we can plug it back
[tex]s(2.040816)=-4.9(2.040816)^2+20(2.040816)+60[/tex]
[tex]s(2.040816)=80.408[/tex]
so, the maximum height is 80.408meter.........Answer
