The height, in feet, of the grasshopper above the ground after t seconds is modeled by the function , which is
[tex] h(t) =-t^2+\frac{4}{3}t+ \frac{1}{4} [/tex]
When the grasshopper land on the ground,
[tex] h(t)=0 [/tex]
that is,
[tex] -t^2+\frac{4}{3}t+ \frac{1}{4}=0 [/tex]
Multiplying whole equation by 12 to get rid of denominators 3 and 4
[tex] -12t^2+16t +3=0 [/tex]
[tex] (3-2t)(6t+1)=0
\\
3-2t=0 , 6t+1=0
\\
t= \frac{3}{2} , \frac{-1}{6} [/tex]
And time cant be negative, so the correct option is
[tex] t = \frac{3}{2}=1.5 seconds [/tex]
Correct option is C .