A ball is thrown from a height of 40 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.
\[
h=40-5 t-5 t^{2}
\]

How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
\[
t=\square \text { seconds } \quad \begin{array}{cc|}
\hline \text { 암 } & \\
\times & 5 \\
\hline
\end{array}
\]

Step 1 ：We are given a ball thrown from a height of 40 meters with an initial downward velocity of 5 m/s. The height of the ball after t seconds is given by the equation \(h=40-5t-5t^{2}\). We need to find the time when the ball hits the ground, i.e., when \(h=0\).

Step 2 ：Setting \(h=0\) in the equation, we get the quadratic equation \(0=40-5t-5t^{2}\). This can be rewritten as \(5t^{2}+5t-40=0\).

Step 3 ：We can solve this quadratic equation using the quadratic formula \(t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), where \(a=-5\), \(b=-5\), and \(c=40\).

Step 4 ：Substituting the values of \(a\), \(b\), and \(c\) into the quadratic formula, we get \(t=\frac{5 \pm \sqrt{(-5)^{2}-4*(-5)*40}}{2*(-5)}\).

Step 5 ：Solving this, we get two solutions for \(t\), \(t1 = -3.37\) seconds and \(t2 = 2.37\) seconds.

Step 6 ：However, time cannot be negative, so we discard the negative solution. Therefore, the ball hits the ground after approximately 2.37 seconds.

Step 7 ：Final Answer: The ball hits the ground after approximately \(\boxed{2.37}\) seconds.