Application of derivatives to motion of a projectile

Let the height of a projectile above the ground be given by

$f(t) = v_0 t - 16t^2$

where $v_0$ is the initial velocity in ft/sec.

Show that the average velocity on the interval $[t, t+h]$ (with $h > 0$ ) is
$v_0 -32t -16h$

and that the velocity at time $t$ is

$v_0 - 32t.$
Find the time at which velocity equals zero.
Find the velocity of the projectile when it gets back to earth (i.e., when f(t) = 0 again).
Find the initial velocity such that the projectile returns to earth in 1 second, 10 seconds, and in $T$ seconds.
Prove that the projectile undergoes constant acceleration.
Find another formula for the height such that the acceleration is a constant -20 ft/sec^2.

The average velocity from time $t$ to time $t+h$ is given by
$\begin{align*} \frac{f(t+h)- f(t)}{h} &= \frac{v_0(t+h) - 16(t+h)^2 - v_0t + 16t^2}{h} \\ &= v_0 - 32t -16h. \end{align*}$

The instantaneous velocity is the limit of this as $h \to 0$ or,

$\begin{align*} \lim_{h \to 0} \frac{v_0(t+h) - 16(t+h)^2 - v_0t +16t^2}{h} &= \lim_{h \to 0} \frac{v_0t + v_0h - 16t^2 - 32th - 16h^2 - v_0t + 16t^2}{h} \\[8pt] &= \lim_{h \to 0} \frac{v_0h - 32th - 16h^2}{h}\\[8pt] &= \lim_{h \to 0} v_0 - 32t - 16h \\ &= v_0 - 32t. \end{align*}$
Let $v_0 -32t = 0$ . This implies $t = \frac{v_0}{32}$ seconds.
We look for the solutions of $f(t) = 0$ .
$f(t) = 0 \quad \implies \quad v_0t - 16t^2 = 0 \quad \implies \quad t = \left\{ 0, \frac{v_0}{16} \right\}.$

The solution $t = 0$ corresponds to the time the projectile was fired. So, the projectile returns to earth at time $t = \frac{v_0}{16}$ , so evaluating $v(t) = f'(t)$ at this time we have,

$v \left( \frac{v_0}{16} \right) = v_0 - 32 \left( \frac{v_0}{16} \right) = -v_0 \text{ ft/sec}.$
If the projectiles returns in 1 second then we have
$f(1) = 0 \quad \implies \quad v_0 - 16 = 0 \quad \implies \quad v_0 = 16 \text{ ft/sec}.$

If the projectile returns in 10 seconds then we have

$f(10) = 0 \quad \implies \quad 10v_0 - 1600 = 0 \quad \implies \quad v_0 = 160 \text{ ft/sec}.$

If the projectile returns in $T$ seconds then we have (with $T > 0$ ),

$f(T) = 0 \quad \implies \quad v_0 T - 16T^2 = 0 \quad \implies \quad v_0 = 16T \text{ ft/sec}.$
Acceleration is given by $f''(t) = \left( f'(t) \right)' = (v_0 -32t)' = -32 \text{ ft/sec}^2$ . This is a constant.
Let $f(t) = v_0 t - 10t^2$ . Then $f'(t) = v_0 - 20t$ and $f''(t) = -20 \text{ ft/sec}^2$ , as requested.