Verified answer

rock .......................... final positionof sound

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l height of the bridge

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final position of rock .... initial position of sound

let t represent the time it takes the rock to hit the water. We know that the total time is 5.62s, that means 5.62 - t, is the time it takes sound to travel back up.

the motion of rock is an acceleration

Xf = .5at^2 + Vt + Xi

Xf = final position (0 m)

acceleration = (-9.8m/s^2)

V = initial position (0 m/s because the rock is dropped)

t = time

Xi = initial position (what we're looking for)

0 = .5(-9.8)t^2 + Xi

the motion of sound is constant.

Xf = Vt + 0

we know the time it takes sound to travel back up is 5.62 - t

Xf = 340(5.62 - t) + 0

Xf = 1910.8 - 340t

so you have two equations

0 = .5(-9.8)t^2 + Xi

Xf (sond) = 1910.8 - 340t

notice take the initial position of the rock is where the sound is heard, that makes Xi (rock) = Xf (sound)

so 0 = .5(-9.8)t^2 - 340t + 1910.8

0 = -4.9t^2 - 340t + 1910.8

use the quadratic formula and you'll get t = 5.22634s

5.62 - 5.22634 = .39366s

so it will takes sound to travel back up in about .39366s

plug the time in to find the heigth of the bridge

Xf = 340(.39366)

Xf =~ 133.8m

hope this helps!