Answer: 3.49 s
Explanation:
We can solve this problem with the following equation of motion:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0 m[/tex] is the final height of the ball
[tex]y_{o}=60 m[/tex] is the initial height of the ball
[tex]V_{o}=0 m/s[/tex] is the initial velocity (the ball was dropped)
[tex]g=9.8 m/s^{2}[/tex] is the acceleratio due gravity
[tex]t[/tex] is the time
Isolating [tex]t[/tex]:
[tex]t=\sqrt{\frac{2 y_{o}}{g}}[/tex] (2)
[tex]t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}}[/tex] (3)
Finally we find the time the ball is in the air:
[tex]t=3.49 s[/tex] (4)