To find the time at which the ball is a certain height above the ground, we need to solve the height equation for $t$ given specific heights. The height equation is:

$H = 20t - 5t^2$

a) When the ball is 15 meters above the ground:

Set $H = 15$ and solve for $t$:

$15 = 20t - 5t^2$

Rearrange the equation:

$5t^2 - 20t + 15 = 0$

Divide the entire equation by 5 to simplify:

$t^2 - 4t + 3 = 0$

Now, solve this quadratic equation using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -4$, and $c = 3$.

$t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

$t = \frac{4 \pm \sqrt{16 - 12}}{2}$

$t = \frac{4 \pm \sqrt{4}}{2}$

$t = \frac{4 \pm 2}{2}$

This gives two possible solutions:

$t = \frac{6}{2} = 3 \quad \text{and} \quad t = \frac{2}{2} = 1$

So, the ball is 15 meters above the ground at $t = 1$ second and $t = 3$ seconds.

b) When the ball is 20 meters above the ground:

Set $H = 20$ and solve for $t$:

$20 = 20t - 5t^2$

Rearrange the equation:

$5t^2 - 20t + 20 = 0$

Divide by 5:

$t^2 - 4t + 4 = 0$

Now solve the quadratic equation:

$t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$t = \frac{4 \pm \sqrt{16 - 16}}{2}$

$t = \frac{4 \pm \sqrt{0}}{2}$

$t = \frac{4}{2}$

$t = 2$

So, the ball is 20 meters above the ground at $t = 2$ seconds.