Answer:
[tex]t=6.4534 s[/tex]
Explanation:
This is an exercise where you need to use the concepts of free fall objects
Our knowable variables are initial high, initial velocity and the acceleration due to gravity:
[tex]y_{0}=75m[/tex]
[tex]v_{oy} =20m/s[/tex]
[tex]g=9.8 m/s^{2}[/tex]
At the end of the motion, the rock hits the ground making the final high y=0m
[tex]y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}[/tex]
If we evaluate the equation:
[tex]0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex]
This is a classic form of Quadratic Formula, we can solve it using:
[tex]t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-4.9\\b=20\\c=75[/tex]
[tex]t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s[/tex]
[tex]t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s[/tex]
Since the time can not be negative, the reasonable answer is
[tex]t=6.4534s[/tex]
