Answer:
This situation is related to parabolic motion and the main equation is:
[tex]y=y_{o}+V_{oy} t-\frac{gt^{2}}{2}[/tex] (1)
Where:
[tex]y=0[/tex] is the final height of the rock, asuming the top of the bridge touches the surface of the water
[tex]y_{o}[/tex] is the initial height of the rock
[tex]V_{oy}=0[/tex] is the vertical component of the initial velocity (it is zero because the rock was thrown horizontally)
[tex]t[/tex] is the time the parabolic motion lasts
[tex]g[/tex] is the acceleration due gravity
Rewritting (1) with these conditions:
[tex]0=y_{o}+(0) t-\frac{gt^{2}}{2}[/tex] (2)
Hence:
[tex]y_{o}=\frac{gt^{2}}{2}[/tex]
