While standing at the edge of a cliff, you throw a stone upward with an initial speed of 6.79 m/s. The stone subsequently falls to the ground, which 12.1 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? Ignore air resistance and take g= 9.81 m/s^2 a) 13.6 m/s
b) 14.5 m/s
c) 15.3 m/s
d) 16.8 m/s

In this situation we are dealing with paarabolic motion with a constant acceleration (acceleration due gravity), hence, the following equation will be useful to find the answer:

[tex]{V_{f}}^{2}={V_{o}}^{2}+2ad[/tex]

Where:

[tex]V_{f}[/tex] Is the final velocity of the stone

[tex]V_{o}=6.79 m/s[/tex] Is the initial velocity of the stone

[tex]a= 9.8 m/s^{2}[/tex] is the constant acceleration of the stone due gravity

[tex]d=12.1 m[/tex] is the height of the cliff

[tex]V_{f}=\sqrt{{V_{o}}^{2}+2ad}[/tex]

[tex]V_{f}=\sqrt{{(6.79 m/s)}^{2}+2(6.79 m/s)(12.1 m)}[/tex]

[tex]V_{f}=16.8 m/s[/tex] This is the final velocity of the stone