help physics

[snowbrood]

Q.1 the survivor of a shipwreck lands on an island which is 300m from a vertical clif he sees a ship anchored between the island and the cliff. a blast from the ship's horn is heard twice with a time lapse of 4s. Calculate the distance d of the ship from the island (Assume that the speed of the sound=330ms).

Q.2 A man stands between two cliffs and claps his hands once he hears two echoes, one after 0.5s and the other 0.3s after the first. if the speed of sound is 330ms what is the distance between the two cliffs?

 

[VelaneDeBeaute]

Alright im gonna solve this question, the second one, but i dunno whether im getting it all the way right !
SO,
Echo 1 = after 0.5 sec
Echo 2 = 0.3 sec after first !
=> Initially we find the distance between the man and cliff 1.
Notice that the echo reaches the man by travelling the distance twice, once going from the man to the cliff, then coming from the cliff back to the man.
So the time is actually twice for the same distance, therein we divide it by 2 and then put it in the formula.
Hence,
d=vt
=330 x 0.5/2
=82.5 m
Next, keep in view that the echo 2 travels the distance between both the cliffs twice as shown by the red lines.
d = vt = 330 x 0.3/2 = 49.5 m.
Reason fer dividing the time is that the sound waves travel the same distance twice !
So total distance b/w both cliffs is gonna be 82.5 + 49.5 = 132 m.

There's another way of doin this though,
it is said that he hears the second echo .3 seconds after the first one so that means the total tym fer the sound wave to travel from man to cliff 1 to cliff 2 back to man is 0.5 + 0.3 seconds = 0.8 sec.
Using the formula,
d = vt = 330 x 0.8 = 264 m
However since the distance is twice, we'd divide it by two & get 132 m as answer !

Hope that was correct !

 

[snowbrood]

Alright im gonna solve this question, the second one, but i dunno whether im getting it all the way right !
SO,
Echo 1 = after 0.5 sec
Echo 2 = 0.3 sec after first !
=> Initially we find the distance between the man and cliff 1.
Notice that the echo reaches the man by travelling the distance twice, once going from the man to the cliff, then coming from the cliff back to the man.
So the time is actually twice for the same distance, therein we divide it by 2 and then put it in the formula.
Hence,
d=vt
=330 x 0.5/2
=82.5 m
Next, keep in view that the echo 2 travels the distance between both the cliffs twice as shown by the red lines.
d = vt = 330 x 0.3/2 = 49.5 m.
Reason fer dividing the time is that the sound waves travel the same distance twice !
So total distance b/w both cliffs is gonna be 82.5 + 49.5 = 132 m.

There's another way of doin this though,
it is said that he hears the second echo .3 seconds after the first one so that means the total tym fer the sound wave to travel from man to cliff 1 to cliff 2 back to man is 0.5 + 0.3 seconds = 0.8 sec.
Using the formula,
d = vt = 330 x 0.8 = 264 m
However since the distance is twice, we'd divide it by two & get 132 m as answer !

Hope that was correct !

hey cant u solve Q.1

 

[VelaneDeBeaute]

I'll try !

 

[Fa7ima]

The above explanation for question 2 is great but you missed out something. The correct answer is 215m. We have to add up 0.5s and 0.3s for the second echo distance.

Formula used: v = 2d/t

Distance for the first echo: [cliff 1 to man]
d=vt
330 x 0.5/2
82.5m

Distance for the second echo: [cliff 2 to man]
d=vt
330 x 0.8/2
132m

Total distance between the two cliffs = 132 + 82.5 = 214.5m


You can get further reference from here: http://books.google.com.pk/books?id=xPPwJNzZyzgC&lpg=PA127&ots=Nv9nUEOUqV&dq=he hears two echoes one after 0.5s&pg=PA127#v=onepage&q=he hears two echoes one after 0.5s&f=false

 

[fatikhalid]

question 1:
i'll try i dont guarantee that the answer is correct!


time=distance/speed
time=x/330


time=distance/speed
time=300-x+300/330

the time difference bw is 4 sec

so,
x/330 - 300-x+300/330 = 4
x= 960m

tell me if i am correct!..