Question

1. an engineer is on top of a large wind turbine replacing an aviation obstruction light. she accidentally drops a screwdriver, which falls to the ground below. it takes 4.0 seconds for the tool to land on the ground.

a. how tall is the wind turbine? (4 pts)
b. what is the velocity of the screwdriver right before it hits the ground? (4 pts)
c. extra credit: sketch the position - time, velocity - time, and acceleration - time graphs for the falling screwdriver. (3 pts)

Explanation:

Step1: Identify the kinematic - equation for height

The screwdriver is in free - fall, and the initial velocity $v_0 = 0$. The kinematic equation for the displacement $y - y_0=v_0t+\frac{1}{2}at^{2}$. Here, $y - y_0$ is the height of the wind turbine $h$, $v_0 = 0$, $a = g=9.8\ m/s^{2}$, and $t = 4.0\ s$. So, $h=\frac{1}{2}gt^{2}$.

Step2: Calculate the height of the wind turbine

Substitute $g = 9.8\ m/s^{2}$ and $t = 4.0\ s$ into the formula $h=\frac{1}{2}gt^{2}$.
\[

$$\begin{align*} h&=\frac{1}{2}\times9.8\times4.0^{2}\\ &=4.9\times16\\ & = 78.4\ m \end{align*}$$

\]

Step3: Identify the kinematic - equation for final velocity

The kinematic equation for final velocity is $v = v_0+at$. Since $v_0 = 0$ and $a = g = 9.8\ m/s^{2}$, $t = 4.0\ s$, then $v=gt$.

Step4: Calculate the final velocity

Substitute $g = 9.8\ m/s^{2}$ and $t = 4.0\ s$ into the formula $v = gt$.
\[

$$\begin{align*} v&=9.8\times4.0\\ &=39.2\ m/s \end{align*}$$

\]

Answer:

a. The height of the wind turbine is $78.4\ m$.
b. The velocity of the screwdriver right before it hits the ground is $39.2\ m/s$.
c.

• Position - time graph: The position - time graph for an object in free - fall with an initial position $y_0$ (top of the turbine) and initial velocity $v_0 = 0$ is a parabola opening downwards. The general form of the position function is $y=y_0-\frac{1}{2}gt^{2}$. If we set the ground as $y = 0$ and the initial position $y_0=h = 78.4\ m$, the function is $y = 78.4-4.9t^{2}$. At $t = 0$, $y = 78.4\ m$; at $t = 4\ s$, $y = 0$.
• Velocity - time graph: The velocity - time graph for an object in free - fall with an initial velocity $v_0 = 0$ is a straight line with a slope of $-g=- 9.8\ m/s^{2}$. The velocity function is $v=-gt$. At $t = 0$, $v = 0$; at $t = 4\ s$, $v=-39.2\ m/s$.
• Acceleration - time graph: The acceleration of an object in free - fall (neglecting air resistance) is constant, $a=-g=-9.8\ m/s^{2}$. So the acceleration - time graph is a horizontal line at $a=-9.8\ m/s^{2}$ for all $t$ during the fall.