A ball is thrown from a height of 49 meters with an initial downward velocity of $2 \mathrm{~m} / \mathrm{s}$. The ball's height $h$ (in meters) after $t$ seconds is given by the following. \[ h=49-2 t-5 t^{2} \] 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.)

Step 1 ：We are given a ball thrown from a height of 49 meters with an initial downward velocity of 2 m/s. The height of the ball after t seconds is given by the equation \(h=49-2t-5t^2\).

Step 2 ：We need to find the time when the ball hits the ground, which is when the height h is zero. So, we need to solve the equation \(49-2t-5t^2 = 0\) for t.

Step 3 ：This is a quadratic equation in the form \(at^2 + bt + c = 0\), where \(a = -5\), \(b = -2\), and \(c = 49\).

Step 4 ：The solutions to this equation are given by 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: \(t1 = -3.3368774282716247\) and \(t2 = 2.9368774282716243\).

Step 6 ：Since time cannot be negative, we discard the negative solution and take \(t = 2.9368774282716243\).

Step 7 ：Rounding to the nearest hundredth, we get \(t = 2.94\) seconds.

Step 8 ：Final Answer: The ball hits the ground approximately \(\boxed{2.94}\) seconds after it is thrown.