a) After 2 seconds the rock will be 30 feet above the surface.
b) After 3.83, 0.17 seconds the rock will be 10 feet above the surface.
a) Given that the height = h(t) =30
Plug it in the equation and find t.
[tex]30=-6t^2+24t+6\\6t^2-24t+24=0\\6(t^2-4t+4)=0\\6(t-2)^2=0\\t=2[/tex]
b) Given that the height = h(t) =10
Plug it in the equation and find t.
[tex]10=-6t^2+24t+6\\6t^2-24t+4=0\\2(3t^2-12t+2)=0\\3t^2-12t+2=0[/tex]
Then use the quadratic formula.
[tex]t=\frac{12 \pm \sqrt{(-12)^2-4(3)(2)} }{2(3)}\\ t=\frac{12 \pm \sqrt{120} }{6}\\\\t=3.83, 0.17[/tex]
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