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anonymous

  • one year ago

PROJECTILE MOTION I don't know how to get started with this. A grasshopper jumps a horizontal distance of 1m from rest, with an initial velocity at a 45 degree angle with respect to the horizontal. Find (a) the initial speed of the grasshopper and (b) the maximum height reached.

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  1. Michele_Laino
    • one year ago
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    here we have to apply the formula of the range \(r\), namely: \[r = \frac{{v_0^2}}{g}\sin \left( {2\theta } \right)\]

  2. Michele_Laino
    • one year ago
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    |dw:1444185191751:dw|

  3. anonymous
    • one year ago
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    Where did you derive that from? Which kinematic equation?

  4. Michele_Laino
    • one year ago
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    that formula expresses the range \(r\) of an object, which is launched with a velocity whose magnitude \(v_0\) and angle (with respect to the horizontal line) equal to \(\theta\)

  5. Michele_Laino
    • one year ago
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    it is a standard formula, which can be derived from the theory of falling bodies

  6. Michele_Laino
    • one year ago
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    in our case we have: \(\theta=45\;degrees\)

  7. Michele_Laino
    • one year ago
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    from my formula above, we get: \[{v_0} = \sqrt {\frac{{rg}}{{\sin \left( {2\theta } \right)}}} = \sqrt {\frac{{1 \cdot 9.81}}{{\sin \left( {2 \cdot 45} \right)}}} = ...m/\sec \]

  8. anonymous
    • one year ago
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    I don't seem to understand that. We're only given three general equations. |dw:1444185562567:dw|

  9. anonymous
    • one year ago
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    I'm just learning Physics by the way. :D

  10. Michele_Laino
    • one year ago
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    yes! we can derive my formula above starting from your formulas, nevertheless you have to keep in mind that the motion of the grasshopper is a composed motion

  11. anonymous
    • one year ago
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    So I'm guessing I'll use the third formula?

  12. Michele_Laino
    • one year ago
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    better is if you follow my replies, furthermore the trajectory of the grasshopper is a parabola, or, in other words, we have a parabolic motion

  13. Michele_Laino
    • one year ago
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    |dw:1444185855561:dw|

  14. anonymous
    • one year ago
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    And we can change \(V_0\) to \(V_0 sin \theta\)? Yeah, I'm sorry if I weren't paying attention to you. I just want to take it slow, I guess, since I'm just beginning to understand it. I just don't want that you're giving me a formula that I've never heard of.

  15. anonymous
    • one year ago
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    Oh that's what I drew, so I guess I can keep up on what you're showing to me.

  16. Michele_Laino
    • one year ago
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    yes! that's right! the derivation of the formula for range \(r\) is long

  17. Michele_Laino
    • one year ago
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    namely it is a long procedure

  18. anonymous
    • one year ago
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    All right. I'll work on that. So what's the next step?

  19. anonymous
    • one year ago
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    3.13 m/s?

  20. Michele_Laino
    • one year ago
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    the maximum height reached by your grasshopper is the vertex of the parabola that I drew: |dw:1444186256072:dw|

  21. Michele_Laino
    • one year ago
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    correct! the initial velocity has a magnitude of \(3.13\) m/sec

  22. anonymous
    • one year ago
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    if it's in the maximum height, then wouldn't \(V_0y=0\)

  23. Michele_Laino
    • one year ago
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    correct! at maximum height the vertical velocity has magnitude equal to zero

  24. Michele_Laino
    • one year ago
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    here is the formula for maximum height: \[\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2}\]

  25. anonymous
    • one year ago
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    0.25 m?

  26. Michele_Laino
    • one year ago
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    substituting our data, we get: \[\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2} = \frac{{{{3.13}^2}}}{{2 \cdot 9.81}}{\left( {\sin 45} \right)^2}=...?\]

  27. Michele_Laino
    • one year ago
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    correct!

  28. anonymous
    • one year ago
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    Oh okay. Thanks!

  29. Michele_Laino
    • one year ago
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    :)

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