Answer: 1.19 s
Step-by-step explanation:
This situation is related to parabolic motion; hence we can use the following equation to find the time the rock is 7 meters from ground level:
[tex]y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}[/tex] (1)
Where:
[tex]y=20m-7m=13 m[/tex] is the height (position) of the rock when it is 7 meters from ground level (taking into account the initial position is 20 m)
[tex]y_{o}=20 m[/tex] is the initial height of the rock
[tex]V_{o}=24 m/s[/tex] is the rock's initial speed
[tex]\theta=0\°[/tex] is the angle at which the rock was thrown (upward or vertically)
[tex]t[/tex] is the time at whch the rock is in position [tex]y[/tex]
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Isolating [tex]t[/tex] from (1):
[tex]t=\sqrt{-\frac{2(y-y_{o})}{g}}[/tex] (2)
Solving with the given data:
[tex]t=\sqrt{-\frac{2(13 m-20 m)}{9.8m/s^{2}}}[/tex] (3)
[tex]t=1.19 s[/tex] (4) This is the time at which the rock is 7 meters from ground level.
