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Math Help - Height of a cliff

  1. #1
    Member Jonboy's Avatar
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    Height of a cliff

    Hello All! I need help with this problem...

    If a rock is is dropped from a cliff into an ocean, it travels approximately 16t^2 feet in t seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

    I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!
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    Quote Originally Posted by Jonboy View Post
    Hello All! I need help with this problem...

    If a rock is is dropped from a cliff into an ocean, it travels approximately 16t^2 feet in t seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

    I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!
    HINT:
    Let s be the distance. Then the time until the sound it heard is equal to the time until the rock hits the water plus the amount of time it takes sound to travel. Use this to set up an equation.
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  3. #3
    Member Jonboy's Avatar
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    Wow ThePerfectHacker that was a extremely expeditious reply!

    So is this correct?

    s\,=\,16(4)^2\,+\,4,400 ?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jonboy View Post
    Hello All! I need help with this problem...

    If a rock is is dropped from a cliff into an ocean, it travels approximately 16t^2 feet in t seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

    I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!
    Quote Originally Posted by ThePerfectHacker View Post
    HINT:
    Let s be the distance. Then the time until the sound it heard is equal to the time until the rock hits the water plus the amount of time it takes sound to travel. Use this to set up an equation.
    Quote Originally Posted by Jonboy View Post
    Wow ThePerfectHacker that was a extremely expeditious reply!

    So is this correct?

    s\,=\,16(4)^2\,+\,4,400 ?
    No.

    It takes the sound 4 seconds to be heard, so let t be the time of fall and T the amount of time it takes for the sound to travel up the cliff. Let the height of the cliff above the water be h.

    Then
    t + T = 4~s

    h = 16t^2

    h = (1100)T

    Solve this system for h.

    -Dan
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    I have something else.

    The amount of time it take for the rock to hit the water is \frac{\sqrt{s}}{4}, now the amount of time for the sound to travel up is \frac{s}{1100}.

    So,
    \frac{\sqrt{s}}{4}+\frac{s}{1100}=4
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  6. #6
    Member Jonboy's Avatar
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    Thanks for the replies. How did you get the following TPH:

    The amount of time it take for the rock to hit the water is , now the amount of time for the sound to travel up is .
    ?
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    Quote Originally Posted by Jonboy View Post
    Thanks for the replies. How did you get the following TPH:
    You should spend more time thinking on these:

    16t^2 = s solve for t.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jonboy View Post
    Thanks for the replies. How did you get the following TPH:



    ?
    Quote Originally Posted by topsquark View Post
    No.

    It takes the sound 4 seconds to be heard, so let t be the time of fall and T the amount of time it takes for the sound to travel up the cliff. Let the height of the cliff above the water be h.

    Then
    t + T = 4~s

    h = 16t^2

    h = (1100)T

    Solve this system for h.

    -Dan
    Question answered. The only difference is I'm using h where you used s.

    -Dan
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  9. #9
    Member Jonboy's Avatar
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    Thank you everyone I understand now!
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  10. #10
    Upendra Bhagat
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    Quote Originally Posted by Jonboy View Post
    Hello All! I need help with this problem...

    If a rock is is dropped from a cliff into an ocean, it travels approximately 16t^2 feet in t seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

    I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!
    According to this Q.,we will get 2 eqns.
    t1=x/16t...........1 , where t1(time taken by the stone to reach sea),16t(Speed of the stone)

    t2=x/1100............2 , where t2(time taken by the sound to reach cliff)


    from 1 and 2,we get t1+t2=x/16t+x/1100......3 and given t1+t2=4 seconds.

    Therefore, eqn 3 changes to
    4=x/16t+x/1100...........4 , which is a 2 variable eqn. so cannot be solved without a extra eqn. or we have to solve it by Hit and Trial or this Q. might have options to solve.
    But it is sure x cannot be equal to 4400.(See the eqn. 4)
    Taking t=1
    , and solving eqn 4, we will get,x=63.08 ft (approx)
    Taking t=2, and solving eqn 4, we will get,x=124.38ft (approx)
    Taking t=3, and solving eqn 4, we will get,x=183.97 ft (approx)

    choose ur ans.


    regards
    upendra
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Upendra Bhagat View Post
    According to this Q.,we will get 2 eqns.
    t1=x/16t...........1 , where t1(time taken by the stone to reach sea),16t(Speed of the stone)
    No because the stone will accelerate downward. This equation assumes the speed of the stone is constant, so it is not physically valid.

    -Dan
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