Question

using the figures from the story above: a student holds a water balloon outside of an open window and lets go. the window is 10 meters above the ground, and the balloon is falling under the acceleration of gravity, which is 9.8 m/s². which of the following equations below provides the correct answer for the distance the balloon has fallen in 1 second? (use the formula sheet that you were provided to help substitute the correct values in this question. note that ^2 means the value prior is squared) (2 points) o x=(0m)+(0 m/s)(1.0 s)+10 (9.8 m/s2)(1s)^2 = 98 m o x=(0m)+(0 m/s)(1.0 s)+½ (9.8 m/s2)(1s)^2 = 4.9 m o v=(10m)+(0 m/s)(1.0 s)+½ (9.8 m/s2)(1s)^2 = 14.9 m o v=(0m)+(0 m/s)(1.0 s)+1 (9.8 m/s2)(1s)^2 = 9.8 m

Explanation:

Step1: Identify the kinematic - equation

The equation for the displacement of an object in free - fall starting from rest ($v_0 = 0$) from an initial position $x_0$ is $x=x_0 + v_0t+\frac{1}{2}at^{2}$, where $x_0$ is the initial position, $v_0$ is the initial velocity, $t$ is the time, and $a$ is the acceleration. Here, $x_0 = 0$ (taking the starting point as the origin), $v_0 = 0$ m/s (released from rest), $a = g=9.8$ m/s², and $t = 1$ s.

Step2: Substitute the values

Substitute $x_0 = 0$ m, $v_0 = 0$ m/s, $a = 9.8$ m/s², and $t = 1$ s into the equation $x=x_0 + v_0t+\frac{1}{2}at^{2}$.
We get $x=(0\text{m})+(0\text{ m/s})(1.0\text{ s})+\frac{1}{2}(9.8\text{ m/s}^2)(1\text{ s})^2$.
Calculate $\frac{1}{2}(9.8\text{ m/s}^2)(1\text{ s})^2=\frac{9.8}{2}\text{ m}=4.9\text{ m}$.

Answer:

$X=(0\text{m})+(0\text{ m/s})(1.0\text{ s})+\frac{1}{2}(9.8\text{ m/s}^2)(1\text{ s})^2 = 4.9\text{ m}$ (the second option)