The phone is in free fall, so it is moving by uniformly accelerated motion, with constant acceleration equal to [tex]g=9.81 m/s^2[/tex] (gravitational acceleration). Taking the water as reference point, the vertical position of the phone at the istant t is
[tex]y(t) = h - \frac{1}{2}gt^2 [/tex]
where h=45.4 m is the height of the bridge. The phone hits the water when y(t)=0, so by requiring this condition we find the time t after which the phone hits the water:
[tex]h- \frac{1}{2}gt^2=0 [/tex]
[tex]h= \frac{1}{2}gt^2 [/tex]
[tex]t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2 \cdot 45.4 m}{9.81 m/s^2} }=3.04 s [/tex]
