Answer
$[v(t)]^2 = v_0^2-19.6~[s(t)-s_0]$
Work Step by Step
$a(t) = \frac{dv}{dt} = -9.8$
$v(t) = \frac{ds}{dt} = v_0-9.8~t$
$s(t) = s_0+v_0~t-4.9~t^2$
We can find an expression for $[v(t)]^2$:
$[v(t)]^2 = (v_0-9.8~t)^2$
$[v(t)]^2 = v_0^2-19.6~v_0~t+(9.8)(9.8)t^2$
$[v(t)]^2 = v_0^2-19.6~v_0~t+(19.6)(4.9)t^2$
$[v(t)]^2 = v_0^2-19.6~[v_0~t-4.9~t^2]$
$[v(t)]^2 = v_0^2-19.6~[s_0+v_0~t-4.9~t^2-s_0]$
$[v(t)]^2 = v_0^2-19.6~[s(t)-s_0]$
