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amishra

  • 3 years ago

A student stands at the top of a cliff and throws a stone horizontally at a speed of 8 m/s. They see the stone splash into the water 5.5 seconds later. a) How high is the cliff? b) What horizontal distance does the stone travel?

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  1. imron07
    • 3 years ago
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    \[t=\sqrt{\frac{2h}{g}}\] You can derive this equation if you want. From here you can find h with known t (a). For b, simply: \[x=v_x*t\]

  2. amishra
    • 3 years ago
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    I'm sorry, can you explain it further? I didn't get it :/

  3. imron07
    • 3 years ago
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    Let's start from beginning: \[h=v_{0y}t+\frac12 gt^2\] Familiar with this?

  4. amishra
    • 3 years ago
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    Yes :) Okay, now what?

  5. imron07
    • 3 years ago
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    since the student throw the stone horizontally, it means v0y=0 can you find h now? :)

  6. amishra
    • 3 years ago
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    g is what?

  7. imron07
    • 3 years ago
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    gravity acceleration, with value 9.8 m/s^2

  8. amishra
    • 3 years ago
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    So, h= (1/2) * 9.8 * (5.5)^2 ?

  9. imron07
    • 3 years ago
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    Good, that's correct.

  10. amishra
    • 3 years ago
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    Oh okay, and how do I find the horizontal distance?

  11. imron07
    • 3 years ago
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    This is the equation for horizontal motion: \[x=v_{0x}t\]

  12. amishra
    • 3 years ago
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    Wait, what is v0x and x? :/

  13. imron07
    • 3 years ago
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    vox is initial horizontal velocity, and x is horizontal displacement.

  14. amishra
    • 3 years ago
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    So, horizontal velocity is 8 m/s? And time is 5.5 seconds?

  15. imron07
    • 3 years ago
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    Yes, now you can find x.

  16. amishra
    • 3 years ago
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    Oh :/ Wow, THANK YOU SO MUCH for spoon feeding me through the entire problem! :D

  17. imron07
    • 3 years ago
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    It's okay. I love teaching :D

  18. amishra
    • 3 years ago
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    And you're damn good at it too :)

  19. imron07
    • 3 years ago
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    Lol, thx for that :)

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