Answer: 125 m
Explanation:
This problem is related to vertical motion and can be solved with the following equation:
[tex]y-y_{o}=V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y-y_{o}=\Delta y[/tex] is the difference between the final height [tex]y[/tex] and initial height [tex]y_{o}[/tex]. Let's take into account the initial height is greater than the final height, since the skydiver is descending.
[tex]V_{o}=0 m/s[/tex] Is the skydiver's initial velocity, assuming the plane was not moving at that moment
[tex]t=5 s[/tex] is the time
[tex]g=10 m/s^{2}[/tex] is the acceleration due gravity
Solving the equation:
[tex]\Delta y=-\frac{1}{2}gt^{2}[/tex] (2)
[tex]\Delta y=-\frac{1}{2}(10 m/s^{2})(5 s)^{2}[/tex] (3)
[tex]\Delta y=-125 m[/tex] (4) Here, the negative sign only indicates the position of the skydiver, and remember the initial height is greater than the final height.
In fact, the distance is positive: 125 m
