Answer
$$
h(t)=15 t-1.86 t^{2}
$$
(a) the velocity of the rock after $2 s$ can be found as follows:
$$
\quad v(t)=h^{\prime}(t)=15-3.72 t .
$$
The velocity after $ 2 s$ is
$$ v(2)=15-3.72(2)=7.56 \mathrm{m} / \mathrm{s}
$$
(b) The velocity of the rock when its height is $25 m$
$$
h(t)=15 t-1.86 t^{2}=25 \\
\Leftrightarrow
1.86 t^{2}-15 t+25=0\\
\Leftrightarrow
t=\frac{15 \pm \sqrt{15^{2}-4(1.86)(25)}}{2(1.86)} \\
\Leftrightarrow t=t_{1} \approx 2.35 \text { or } t=t_{2} \approx 5.71
$$
The velocities are:
$$v\left(t_{1}\right)=15-3.72 t_{1}=15-3.72 (2.35)\approx 6.24 \mathrm{m} / \mathrm{s}
[\text { upward }] $$
and
$$
v\left(t_{2}\right)=15-3.72 t_{2}=15-3.72 (5.71) \approx-6.24 \mathrm{m} / \mathrm{s}[\text { downward }]$$
Work Step by Step
$$
h(t)=15 t-1.86 t^{2}
$$
(a) the velocity of the rock after $2 s$ can be found as follows:
$$
\quad v(t)=h^{\prime}(t)=15-3.72 t .
$$
The velocity after $ 2 s$ is
$$ v(2)=15-3.72(2)=7.56 \mathrm{m} / \mathrm{s}
$$
(b) The velocity of the rock when its height is $25 m$
$$
h(t)=15 t-1.86 t^{2}=25 \\
\Leftrightarrow
1.86 t^{2}-15 t+25=0\\
\Leftrightarrow
t=\frac{15 \pm \sqrt{15^{2}-4(1.86)(25)}}{2(1.86)} \\
\Leftrightarrow t=t_{1} \approx 2.35 \text { or } t=t_{2} \approx 5.71
$$
The velocities are:
$$v\left(t_{1}\right)=15-3.72 t_{1}=15-3.72 (2.35)\approx 6.24 \mathrm{m} / \mathrm{s}
[\text { upward }] $$
and
$$
v\left(t_{2}\right)=15-3.72 t_{2}=15-3.72 (5.71) \approx-6.24 \mathrm{m} / \mathrm{s}[\text { downward }]$$
