I'll post the question here and then show my workings out.
A body is thrown into the air with an initial velocity of $v_0$ ft/sec. What initial velocity is required to double the maximum height previously attained?
First I write the equation for velocity in feet.
$v=-32t+v_0$
Integrate to get s.
$s=-16t^2 + v_0t$
Calculate the maximum.
when $v=0$, $\frac{v_0}{32}=t$
Twice the distance is $2s=-32t^2+2v_0 t$
Substituting $\frac{v_0}{32}=t$ into this equation gives $2s=\frac{-v_0^2}{32}+\frac{v_0}{4}$
I then differentiate $2s$, now as $s_2$ to get the velocity required in terms of initial velocity.
$v_2=\frac{4-v_0}{16}$ $v_2=\frac{2-\sqrt(v_0)}{4}$
This is not the correct answer, suggestions would be greatly appreciated!
Recall that
$${v_f}^2-{v_i}^2=2a\Delta x$$
where $v_i$ and $v_f$ denote initial and final velocities, respectively; $a$ is acceleration; and $\Delta x$ is the net displacement.
At its maximum height, the object has $0$ vertical velocity, so if it is thrown with initial velocity $v_0$ and is in freefall, then
$$0^2-{v_0}^2=-2g\Delta x\implies \Delta x=\dfrac{{v_0}^2}{2g}$$
where $g$ is the magnitude of the acceleration due to gravity.
If one wants the object to reach a maximum height twice as high, that would require scaling up $v_0$ by a factor of $\sqrt2$, since
$$\dfrac{(\sqrt2\,v_0)^2}{2g}=\dfrac{{v_0}^2}g=2\Delta x$$