The object will hit the ground in 4.3 s.
The formula used for the object thrown by Chay is:
[tex]h(t)=-16t^{2}+v_{0}t+h_{0}[/tex]
According to the problem,
Initial velocity [tex]v_{0}=60ft/s[/tex]
Initial height [tex]h_{0}=40ft[/tex]
Therefore,
[tex]h(t)=-16t^{2}+60t+40[/tex]
When the object hits the ground h(t) becomes 0.
Thus,
[tex]0=-16t^{2}+60t+40[/tex]
Dividing both sides by 4 we get;
[tex]4t^{2}-15t-10=0[/tex]
Solving by applying quadratic formula;
[tex]x=\frac{-b+/-\sqrt{b^{2-4ac} } }{2a}[/tex]
[tex]t=\frac{15+/-\sqrt{15^{2}+160 } }{8}[/tex]
t=4.3s
