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.
$$h=4+29t-16t^2$$
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 $a{t}^2+bt+c=0$. 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 $t=\boxed{{\displaystyle 0.64}}$ seconds or $t=\boxed{{\displaystyle 1.17}}$ seconds.