Answer
(a) $19\,\mathrm{m/s}$
(b) $9.03\,\mathrm{m/s^2}$
(c) $2.1\,\mathrm{s}$
(d) $8.94\,\mathrm{m/s}$
Work Step by Step
(a) For the motion on the incline:
$v_0=0$, $a=4.00\,\mathrm{m/s^2}$ and $\Delta x=45.0\,\mathrm{m}$
Using
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
&=0+2(4.00)(45.0)\\
&=360\\
v&=\sqrt{360}\\
v&=19\,\mathrm{m/s}
\end{align*}
(b) For motion in the horizontal pool:
$v_0=19\,\mathrm{m/s}$, $v=0$ and $\Delta x=20.0\,\mathrm{m}$.
To compute the deceleration:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
0&=19^2+2a(20.0)\\
a&={-361\over 40.0}\\
a&=-9.03\,\mathrm{m/s^2}
\end{align*}
Hence the deceleration in the pool is $9.03\,\mathrm{m/s^2}$
(c) To compute the time to stop, we have:
$v=0$, $v_0=19\,\mathrm{m/s}$ and $a=-9.03\,\mathrm{m/s^2}$.
Using the kinematic equation:
\begin{align*}
v&=v_0+at\\
0&=19+(-9.03)t\\
t&={19\over 9.03}\\
t&=2.1\,\mathrm{s}
\end{align*}
It takes $2.1\,\mathrm{s}$ to stop.
(d) The speed at the end of the first $10.0\,\mathrm{m}$:
\begin{align*}
v^2=v_0^2+2a\Delta x\\
&=0+2(4.00)(40.0)\\
&=80.0\\
v&=\sqrt{80.0}\\
v&=8.94\,\mathrm{m/s}
\end{align*}
