一个球在一跳直线上滚动，它与起点的距离 $s\text{ cm}$ 可以用以下方程建模：
$s = at + \frac{b}{\sqrt{t} + 4} + c,$
其中 $t$ 是以秒为单位的时间，$a$、$b$ 和 $c$ 是实常数。 当 $t = 0$ 时，球在起点，并且在最初的 5 秒内移动了 10 cm。在接下来的 16 秒内它又移动了 9 cm。求当 $t = 50$ 时球与起点的距离。

1. Use the initial condition at $t = 0$ to express $c$ in terms of $b$.

When $t = 0$, the ball is at the starting point, meaning $s = 0$.
$0 = a(0) + \frac{b}{\sqrt{0} + 4} + c$
$0 = 0 + 0.25b + c \implies c = -0.25b$

2. Use the condition for the first 5 seconds to form an equation relating $a$ and $b$.

When $t = 5$, the distance $s = 10$.
$10 = 5a + \frac{b}{\sqrt{5} + 4} + c$
Substituting $c = -0.25b$:
$10 = 5a + b\left(\frac{1}{\sqrt{5} + 4} - 0.25\right)$

3. Use the condition for the next 16 seconds to form a second equation relating $a$ and $b$.

The ball moved another $9$ cm in the next $16$ seconds, so at $t = 5 + 16 = 21$, the distance from the start is $s = 10 + 9 = 19$.
$19 = 21a + \frac{b}{\sqrt{21} + 4} + c$
Substituting $c = -0.25b$:
$19 = 21a + b\left(\frac{1}{\sqrt{21} + 4} - 0.25\right)$

4. Solve the simultaneous equations to find the values of $a$, $b$, and $c$.

Equating (1) and (2):
$2 + \frac{\sqrt{5}}{20(\sqrt{5} + 4)}b = \frac{19}{21} + \frac{\sqrt{21}}{84(\sqrt{21} + 4)}b$
$b \left( \frac{\sqrt{5}}{20(\sqrt{5} + 4)} - \frac{\sqrt{21}}{84(\sqrt{21} + 4)} \right) = \frac{19}{21} - 2$
Using calculator values to at least 5 significant figures for intermediate accuracy:
$b (0.017928 - 0.0063564) \approx -1.0952$
$0.011572 b \approx -1.0952 \implies b \approx -94.645$
Substitute $b$ back to find $a$ and $c$:
$c = -0.25(-94.645) \approx 23.661$
$a \approx 2 + 0.017928(-94.645) \approx 0.30316$

5. Calculate the distance $s$ when $t = 50$.

Substitute $a$, $b$, $c$ and $t = 50$ into the model:
$s = 50(0.30316) + \frac{-94.645}{\sqrt{50} + 4} + 23.661$
$s \approx 15.158 - 8.549 + 23.661 = 30.270$
Rounding to 3 significant figures, $s \approx 30.3$ cm.