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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.

  1. 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

        \[ v_0 - 32t. \]

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

  1. 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*}

  2. Let v_0 -32t = 0. This implies t = \frac{v_0}{32} seconds.
  3. 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,

        \[ v \left( \frac{v_0}{16} \right) = v_0 - 32 \left( \frac{v_0}{16} \right) = -v_0 \text{ ft/sec}. \]

  4. 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),

        \[ f(T) = 0 \quad \implies \quad v_0 T - 16T^2 = 0 \quad \implies \quad v_0 = 16T \text{ ft/sec}. \]

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

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