where h(t) is the height of the watermelon, and t is time in seconds
The function can also be expressed as :
[tex]h(t)=-4(2t+3)(2t-5)[/tex]when the watermelon hits the ground, h(t) = 0, then:
[tex]\begin{gathered} 0=-4(2t+3)(2t-5) \\ 0=(2t+3)(2t-5) \end{gathered}[/tex]we have two options:
[tex]\begin{gathered} 2t+3=0 \\ 2t=-3 \\ t=-\frac{3}{2} \end{gathered}[/tex]or
[tex]\begin{gathered} 2t-5=0 \\ 2t=5 \\ t=\frac{5}{2} \\ t=2.5 \end{gathered}[/tex]The negative solution has no sense in the context of this problem, then the watermelon takes 2.5 seconds to hit the ground