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arcticf0x

A particle during projectile motion reaches height h in time t1. Again it reaches this height 'h' in time t2 measured from start. Show that the height of point 'h' is 1/2 gt1t2.

  • 2 years ago
  • 2 years ago

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  1. y2o2
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    what does it mean "from start" ?!

    • 2 years ago
  2. arcticf0x
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    from initial point

    • 2 years ago
  3. JamesJ
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    What's the equation of motion of the projectile? Let y be its height above the point from which it's thrown. Then \[ y(t) = v_0t - \frac{g}{2}t^2 \] Now you're told that for two values of time, \( t = t_1, t_2 \) \[ y(t_1) = y(t_2) = h \] Use these equations to solve for h.

    • 2 years ago
  4. arcticf0x
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    Already did that, the problem is, how to merge t1, t2. And the equation of projectile is \[x \tan \alpha - gx ^{2}/2u ^{2}\cos ^{2}\alpha\]

    • 2 years ago
  5. arcticf0x
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    thats the y component

    • 2 years ago
  6. JamesJ
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    The equation I've written down is just the vertical component, the y. And that's all you need. The x component isn't necessary.

    • 2 years ago
  7. JamesJ
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    Substitute t = t1 and t2 into that equation and you have two simultaneous equations in the variables v_0 and h. Now solve them for h.

    • 2 years ago
  8. JamesJ
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    I.e., \[ \frac{g}{2}t_1^2 + v_0 t_1 = h \] \[ \frac{g}{2}t_2^2 + v_0 t_2 = h \] Solve for h.

    • 2 years ago
  9. JamesJ
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    *dropped the minus sign on the g terms.

    • 2 years ago
  10. arcticf0x
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    yeah i am almost there now. I eliminated v0 and the one equation is left in the terms i need. Its not rearraning easily but i will leave it to that because i have got the idea. Thanks again JamesJ!

    • 2 years ago
  11. JamesJ
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    Multiply the first equation by t_2; the second by t_1. Then subtract the second equation from the first and the v_0 terms will cancel.

    • 2 years ago
  12. arcticf0x
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    You are a genius and a lifesaver, just cannot thank you enough! It worked :)

    • 2 years ago
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