$v_{0}\mbox{ \(m/sec\)}$. We learn in physics that the height of the rock above the ground at time $t$ after it left the platform is $$h(t) = H + v_{0}t - ct^{2} \label{eq:vert}$$ where $c$ is a constant. Find the velocity function $v(t)$ of the rock. (Hint: use the method of the previous subsection. Make sure that you treat “h” and “H” as two different symbols.)\
(b) Find an expression for $t_{m}$, the time when the velocity function equals zero.\
(c) Find an expression for the slope of $h(t)$ at time $t_{m}$.\
(d) Find an expression for the straight line that is tangent at time $t_{m}$ to the curve described by $h(t)$ on a distance-time graph.\
(e) Find an expression for the maximum value of $h(t)$ as a function of $t$. (Hint: to check your work, the answer to this part, when $H = 8\mbox{ \(m\); }$ $v_{0}= 12\mbox{ \(m/sec\);}$ and $c = 5\mbox{ \(m/sec^{2}\)}$, is $15.2\mbox{ \(m\)}$.\
(f) Make a graph of $h(t)$ on the vertical axis against $t$ on the horizontal axis, using the values in part (e). The horizontal axis ofyour graph should be at least 15 cm long, and the horizontal scale should range from 0 to 