a rock is drop from a bridge, an observer hears it fall 20 seconds
later, taking the speed of sound into consideration how high is the
bridge?

---

Request for Question Clarification by maniac-ga on 17 Jan 2004 13:04 PST

Hello Cin1,

I have an answer to your question assuming no drag where the speed at
impact is about 1/2 the speed of sound. Is that a good assumption or
are there some other factors to take into consideration. For example,
are specific values for acceleration (9.78 m/s^2) and the speed of
sound (340.29 m/s) that you need to use?

  --Maniac

You can't answer this question without knowing how far the observer is
from the place where the rock hits...

Hi..

hailstorm is right..

assuming acceleration of gravity = 9.78 m/s and speed of sound 340.29 m/s^2
and no drag force:

we get
(all equations in the simplest form)

s1  = 0.5 (9.78) * (t1)^2 -- (1)
s2  = 340.29 * t2         -- (2)
t1 + t2 = 20              -- (3)

where 
s1 = height of bridge ; 
s2 = distance from where the rock hit the ground to the listener ; 
t1 = time from when the rock is released to when it hit the ground ;  
t2 = time from when the rock hit the ground to when the listener hear the sound.

so, you have 4 unknown and only 3 equations.. which makes it impossible to get
the exact answer.

the best you can do is plotting the graph of s1 based on assumed value of
t1 ranging from 0 to 20 seconds..

hope that helped..

-Hiko-