A ball is thrown from an initial height of 4 feet with an initial upward velocity of $29 \mathrm{ft} / \mathrm{s}$. The ball's height $h$ (in feet) after $t$ seconds is given by the following.
\[
h=4+29 t-16 t^{2}
\]
Find all values of $t$ for which the ball's height is 16 feet.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Step 1 ：Given the equation for the height of the ball is \(h=4+29t-16t^2\), we need to find the time \(t\) when the height \(h\) is 16 feet. This means we need to solve the equation \(4+29t-16t^2=16\) for \(t\).

Step 2 ：This is a quadratic equation, and we can solve it using the quadratic formula \(t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation \(at^2+bt+c=0\). In this case, \(a=-16\), \(b=29\), and \(c=4-16=-12\).

Step 3 ：Substituting the values of \(a\), \(b\), and \(c\) into the quadratic formula, we get two solutions for \(t\), which are \(t1 = 1.17\) and \(t2 = 0.64\).

Step 4 ：Final Answer: The ball is at a height of 16 feet at times \(t=\boxed{0.64}\) seconds or \(t=\boxed{1.17}\) seconds.