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catsrule332

  • 2 years ago

A rock is pushed off of a cliff. If air resistance can be neglected, how far will the rock have fallen after 4s? Explain.

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  1. loophole144
    • 2 years ago
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    acceleration due to gravity=9.8 m/s^2

  2. Callum29
    • 2 years ago
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    If the rock is pushed off (supposing the push is in the horizontal direction so it does not affect the motion) then the initial vertical velocity is 0ms^-1. gravity is 9.8ms^-2 and we have that: \[S= ut + at^2 = 0 + 9.8(4)^2 = 156.8m\]

  3. catsrule332
    • 2 years ago
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    I understand that but I'm not sure I set the equation up right. I used: \[DeltaX=1/2at ^{2}\]

  4. Callum29
    • 2 years ago
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    sorry, \[S= ut + \frac{1}{2}at^2 = 4.9\times 16 = 78.4m\]

  5. catsrule332
    • 2 years ago
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    Okay so the 1/2 is right?

  6. Callum29
    • 2 years ago
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    that is the formula one should use...

  7. Callum29
    • 2 years ago
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    Yeah, the 1/2 is correct there in that formula. \[S= ut + \frac{1}{2}at^2\] Where S is the distance, u is the initial velocity, t is time and a is acceleration.

  8. Callum29
    • 2 years ago
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    I remember I proved it once but can't remember the other formulae I used in the proof, it's just a substitution thing anyway

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