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## Homework Statement

From Tipler & Mosca, Ch. 2, #56 (pg. 57)

Bats use echo location to place themselves. (I condensed all of this from the problem) So:

Bat flying at 19.5 m/s toward a vertical cliff.

Bat clicks.

Bat hears the echo 0.15 s later. (click travels at 343 m/s)

How close is the bat when it hears the echo?

## Homework Equations

The position function for the bat is straightforward: [itex]X_{bat}=19.5t[/itex]

the position function for the sound takes a little reasoning. The sound has to travel to the wall and back, so I'm going with [itex]X_{click}={1/2} * 343t[/itex].

## The Attempt at a Solution

Solve these functions for t=0.15:

[itex]X_{f bat}=19.5 m/s(0.15 s)=2.925 m[/itex]

[itex]X_{click}={1/2} * 343 m/s(0.15 s)=25.725 m=X_{i bat}[/itex]

So, the bat's initial position is 25.725 m from the wall, and it's final position would be:

[itex]X_{f bat}= X_{i bat} - X_{f bat} = 25.725 m - 2.925 m = 22.8 m[/itex]

So, the bat is 22.8 m from the wall when it hears the click.