Since this is a free fall and all the motion is going down, let's make down the positive direction (that way we won't use minus signs). A free fall is an uniform accelerated motion, this means that we can use the formula:
[tex]y-y_0=v_0t+\frac{1}{2}at^2[/tex]In this case we know the initial velocity is 10 m/s, the time is 2.1 and the acceleration of gravity is 9.8 m/s^2. Plugging this values we have that:
[tex]\begin{gathered} y-y_0=10(2.1)+\frac{1}{2}(9.8)(2.1)^2 \\ y-y_0=42.609 \end{gathered}[/tex]This means that the rock fell 42.609 meters. Therefore, the bridge is 42.609 m tall.