Answer: 6.45 s
Explanation:
We have the following equation:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0[/tex] is the height when the rock hits the ground
[tex]y_{o}=75 m[/tex] the height at the edge of the cilff
[tex]V_{o}=20 m/s[/tex] the initial velocity
[tex]g=9.8 m/s^{2}[/tex] acceleration due gravity
[tex]t[/tex] time
[tex]0=75 m+(20 m/s)t-(4.9 m/s^{2})t^{2}[/tex] (2)
Rearranging the equation:
[tex]-(4.9 m/s^{2})t^{2} + (20 m/s)t + 75 m=0[/tex] (3)
At this point we have a quadratic equation of the form [tex]at^{2}+bt+c=0[/tex], and we have to use the quadratic formula if we want to find [tex]t[/tex]:
[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex] (4)
Where [tex]a=-4.9[/tex], [tex]b=20[/tex], [tex]c=75[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]t=\frac{-20\pm\sqrt{20^{2}-4(-4.9)(75)}}{2(-4.9)}[/tex] (5)
[tex]t=6.453 s[/tex] This is the time it takes to the rock to hit the ground
