Answer:
About 4.1323 meters.
Step-by-step explanation:
We can use the following kinematic equations:
[tex]\displaystyle v_f = v_0 + at \text{ and } \Delta d = v_0 t + \frac{1}{2} at^2[/tex]
We want to determine the distance the object has dropped after falling from rest and reaching an instantenous speed of 9 m/s.
We are given that the acceleration due to gravity is 9.8 m/s².
Using this fact and the first equation, find the time for which it took the object to reach 9 m/s. Note that the initial velocity is 0 m/s since the object started from rest.
[tex]\displaystyle \begin{aligned} v_f = v_0 + at \\ \\ (9\text{ m/s}) & = (0\text{ m/s}) + (9.8\text{ m/s$^2$})t \\ \\ 9\text{ m/s} & = (9.8 \text{ m/s$^2$})t \\ \\ t &\approx0.9184 \text{ s} \end{aligned}[/tex]
To find how far the object dropped, we can use the second equation:
[tex]\displaystyle\begin{aligned} \Delta d & = v_0 t + \frac{1}{2} at^2 \\ \\ & = (0\text{ m/s})(0.9184 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s$^2$})(0.9184 \text{ s})^2 \\ \\ & = 4.1323 \text{ m} \end{aligned}[/tex]
In conclusion, the object would have dropped about 4.1323 meters.
