Answer:
t= +/- [tex]\frac{3\sqrt{6}}{4}[/tex]
Step-by-step explanation:
set h=0, 0=-16[tex]t^{2}[/tex]+54, add the -16[tex]t^{2}[/tex] to the other side, 16[tex]t^{2}[/tex]=54, divide the 16 off, [tex]t^{2}[/tex]=[tex]\frac{54}{16}[/tex], square root both sides, t= +/- [tex]\sqrt[]{\frac{54}{16} }[/tex] , t= +/- [tex]\sqrt{54}[/tex]/4, t= +/- [tex]\frac{3\sqrt{6}}{4}[/tex] , the object will hit the ground when t= +/- [tex]\frac{3\sqrt{6}}{4}[/tex]
