A ball is thrown from a height of 20 meters with an initial downward velocity of $5 \mathrm{~m} / \mathrm{s}$. The ball's height $h$ (in meters) after $t$ seconds is given by the following.
\[
h=20-5 t-5 t^{2}
\]

How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.

Step 1 ：A ball is thrown from a height of 20 meters with an initial downward velocity of 5 m/s. The ball's height h (in meters) after t seconds is given by the equation \(h=20-5t-5t^2\).

Step 2 ：We need to find out when the ball hits the ground, which is when its height is 0. So, we need to solve the equation \(20-5t-5t^2 = 0\) for t.

Step 3 ：This is a quadratic equation in the form \(at^2 + bt + c = 0\), where \(a = -5\), \(b = -5\), and \(c = 20\).

Step 4 ：We can solve this equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 5 ：Substituting the values of a, b, and c into the quadratic formula, we get two solutions for t: \(t1 = -2.56\) and \(t2 = 1.56\).

Step 6 ：However, time cannot be negative, so we discard the negative solution.

Step 7 ：Therefore, the ball hits the ground after approximately 1.56 seconds.

Step 8 ：Final Answer: The ball hits the ground after approximately \(\boxed{1.56}\) seconds.