You attach a meter stick to an oak tree, such that the top of the meter stick is 2.07 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.306 seconds to pass the length of the meter stick, how high họ above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the way down? ho = m * TOOLS x10

The diagram of the system is given below:

According to the Newton's law of motion , the vertical distance covered by an object initially at rest is calculated by using the formula $h=\frac{1}{2}g{t}^2$ , where $g=9.8 m/s^2$ is the acceleration due to gravity and t is the time taken.

• When the acorn reaches the top of the scale , the formula will be $h_0-2.07 m=\frac{1}{2}g{t_1}^2 \to \left(1\right)$ , where $h_0$ is the initial height of the acorn and $t_1$ is the time taken to reach the height $h=2.07 m$ .
• When the acorn reaches the bottom of the scale , the formula will be $h_0-1.07 m=\frac{1}{2}g{t_2}^2 \to \left(2\right)$ , where $h_0$ is the initial height of the acorn and $t_2$ is the time taken to reach the height $h\text{'}=1.07 m$ , where the time taken to cross the meter scale is given as $t_2-t_1=0.306 s$.

Substract the equation (1) from (2) , thus ,

$\begin{array}{rcl}\left(h_0-2.07 m\right)-\left(h_0-1.07 m\right)& =& \frac{1}{2}g\left({t_1}^2-{t_2}^2\right)\\ -1m& =& \frac{1}{2}\times \left(9.8 m/s^2\right)\times \left({t_1}^2-{t_2}^2\right)\\ 1 m& =& \left(4.9 m/s^2\right)\times \left({t_2}^2-{t_1}^2\right)\\ \left(t_2-t_1\right)\left(t_2+t_1\right)& \approx & 0.204 s^2\\ t_2+t_1& \approx & \frac{0.204 s^2}{0.306 s}\\ & \approx & 0.7 s\end{array}$