I thought that I could do it like this. Given that $$g=9.8m/s^2$$
$$\int -9.8 \, dt=v_0-9.8 t$$
Setting it equal to zero we have: $$t=\frac{v_0}{9.8}$$
$$\int \left(v_0-9.8 t\right) \, dt=-4.9 t^2+t v_0+y_0$$
Then I substitute
$$0.0510204 v_0^2+y_0=h$$,
Now I do not understand how the answer is $$4 h m$$ and $$\sqrt{2} v_0 m/s$$
On
Remember that by chain rule, you have $a = \frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = v\frac{dv}{ds}$. Now separate, integrate (remember to get the limits right, since you're going up, but acceleration is downwards). Now you should get an analytical expression which should allow exploring what happens when you change $v$ or $h$.
Also, I would suggest not plugging any values (for example $g$) into the equation unless necessary. That keeps it less cluttered.
On
This may seem reasonable at first glance $$\int -9.8 \, dt=v_0-9.8 t$$ But try to reconsider. The question is asking how high it would have gone if the ball's initial velocity were $2v_0$. Thus the result of integration should be $$\int -9.8 \, dt=2v_0-9.8 t$$ This should give you the correct $t$ for the maximum height
With the initial velocity of $v_{0}$, the velocity of the traveling ball is given by the equation $$v(t) = v_{0} - gt.$$ At the maximum height of $h$ (where it reaches after some time $t_{m}$), $v(t_{m}) = 0$. So setting $v(t_{m}) = 0$ in the equation, we know it reaches the max height at time $t_{m} = v_{0}/g$. The max height is $$h = \int_{0}^{t_{m}} v(t) \ dt = v_{0}^{2}/2g.$$
Now if the initial velocity is $2v_{0}$, then the equation becomes $$v(t) = 2v_{0} - gt.$$ So it takes $t_{m} = 2 v_{0}/g$ to reach its max height. The max height is $$h = \int_{0}^{t_m} v(t) \ dt = 2v_{0}^{2}/g,$$ which is $4$ times the max height if it were instead thrown with initial velocity of $v_{0}$.