Answer:
About 0.2041 meters.
Explanation:
We can use the following kinematic equation to determine the time in which the grape was in the air. (Ignoring air resistance.)
[tex]\displaystyle v_f = v_i + at[/tex]
The final velocity is 0 m/s (the velocity at the peak is always zero), the initial velocity is 2 m/s, and the acceleration (due to gravity) is -9.8 m/s². Solve for time t:
[tex]\displaystyle \begin{aligned} (0\text{ m/s}) & = (2 \text{ m/s}) + (-9.8 \text{ m/s$^2$})t \\ \\ t & =\frac{-2\text{ m/s}}{-9.8 \text{ m/s$^2$}} \\ \\ & =0.2041 \text{ s} \end{aligned}[/tex]
To find the distance traveled, we can another kinematic equation:
[tex]\displaystyle \Delta d = v_i t + \frac{1}{2} a t^2[/tex]
The initial velocity is 2 m/s, the acceleration (due to gravity) is -9/8 m/s², and the time it took for the graph to reach its maximum height is 0.2041 seconds. Hence:
[tex]\displaystyle \begin{aligned} \Delta d & = (2\text{ m/s})(0.2041 \text{ s}) + \frac{1}{2}(-9.8 \text{ m/s$^2$})(0.2041\text{ s})^2 \\ \\ & = 0.2041 \text{ m} \end{aligned}[/tex]
In conclusion, the grape travels 0.2041 meters in the air.
