we can use height formula
[tex]h(t)=-\frac{1}{2} g t^2+vt+h(0)[/tex]
where
h(t) is the height after t seconds
v is velocity
h(0) is initial height
g is acceleration due to gravity
we know that
g=32 ft/s^2
we have
v=18.5ft/s
h(0)=4.5ft
so, we can plug these values
and we get
[tex]h(t)=-\frac{1}{2}*32 t^2+18.5t+4.5[/tex]
[tex]h(t)=-16 t^2+18.5t+4.5[/tex]
now, we have to find time when object will hit the ground
and we know that object will hit the ground when height becomes 0
so, we can set h(t)=0
and then we can solve for t
[tex]0=-16 t^2+18.5t+4.5[/tex]
we can use pythagoras theorem
[tex]t=\frac{-185-\sqrt{185^2-4\left(-160\right)45}}{2\left(-160\right)}=\frac{37+\sqrt{2521}}{64}[/tex]
[tex]t=1.36265sec[/tex]..............Answer
