Answer
\begin{align}
&(a) \ t_{max} = 4.9s \\
&(b) \ s_{max} = 270m \\
&(c) \ t = 9.85s \\
& (d) \ v = -49m/s \\
& (e) \ t_{total} = 12.74s \\
& (f) \ v_{ground} = -77.9m/s
\end{align}
Work Step by Step
$\text {It is given that the initial velocity and height:}$
\begin{align}
v_0=49 \ m/s; \ s_0 = 150 \ m
\end{align}
$\text {In order to solve this problem, we have to use formulas (15) and (16) from book:}$
\begin{align}
& s(t) = s_0+v_0t-\frac{gt^2}{2} \\
& v(t) = v_0 -gt
\end{align}
$\text {Also, we know that g $\approx$ 10 m/$s^2$. Thus,}$
$\text {(a) At maximum height v = 0:}$
\begin{align}
& v(t) = v_0 -gt \\
& 0 = 49 - 10\times t_{max} \\
&t_{max} = 4.9s
\end{align}
$\text {(b)}$
\begin{align}
& s(t) = s_0+v_0t_{max}-\frac{gt_{max}^2}{2} \\
& s_{max} = 150 + 49 \times 4.9-\frac{10 \times 4.9^2}{2} = 270m
\end{align}
$\text {(c)}$
\begin{align}
t = t_{down1} + t_{max} = 2 \times t_{max} = 9.8s
\end{align}
$\text {(d) At maximum height v = 0:}$
\begin{align}
& v(t) = v_0 -gt_{down1} \\
& v = 0 - 10\times 4.9 = -49m/s
\end{align}
$\text {(e) At the ground s = 0:}$
\begin{align}
s(t) = s_0+&vt_{down2}-\frac{gt_{down2}^2}{2} \\
0 = 150 - &49t_{down2} - 5t_{down2}^2 \\
&t_{down2} = 2.89s \\
t_{total} &= t + t_{down2} = 12.74s
\end{align}
$\text {(f)}$
\begin{align}
&v(t) = v_0-gt_{down2} \\
& v_{ground} = -49 -10 \times 2.89 = -77.9 m/s
\end{align}
