Question

An object falls from a position of rest and

Answer 1

Answer:

About 4.1323 meters.

Step-by-step explanation:

We can use the following kinematic equations:

$\displaystyle v_f = v_0 + at \text{ and } \Delta d = v_0 t + \frac{1}{2} at^2$

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.

$\displaystyle \begin{aligned} v_f = v_0 + at \\ \\ (9\text{ m/s}) & = (0\text{ m/s}) + (9.8\text{ m/s ^2 })t \\ \\ 9\text{ m/s} & = (9.8 \text{ m/s ^2 })t \\ \\ t &\approx0.9184 \text{ s} \end{aligned}$

To find how far the object dropped, we can use the second equation:

$\displaystyle\begin{aligned} \Delta d & = v_0 t + \frac{1}{2} at^2 \\ \\ & = (0\text{ m/s})(0.9184 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s ^2 })(0.9184 \text{ s})^2 \\ \\ & = 4.1323 \text{ m} \end{aligned}$

In conclusion, the object would have dropped about 4.1323 meters.