The function h is given by:
[tex]\begin{gathered} h(t)=-16t^2+v_0t+h_0 \\ \text{ Substitute }v_0=96\text{ and }h_0=640\text{ into the equation:} \\ \end{gathered}[/tex]
Therefore,
[tex]h(t)=-16t^2+96t+640[/tex]
When the object reaches the ground, h(t)=0:
[tex]-16t^2+96t+640=0[/tex]
Divide both sides by -16:
[tex]\begin{gathered} t^2-6t-40=0 \\ t^2+4t-10t-40=0 \\ t(t+4)-10(t+4)=0 \\ (t-10)(t+4)=0 \\ \text{ Therefore,} \\ t=10,-4 \end{gathered}[/tex]
Since t is not negative, discard the negative value of t.
Therefore, the required time is t = 10s
