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anonymous

  • 5 years ago

show that flight time can be estimated by t=sqrt ((2h-2Rtan a)/g)

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  1. anonymous
    • 5 years ago
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    What's the question that leads to this?

  2. anonymous
    • 5 years ago
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    There isn't one but the information that i have is that the velocity vector is <vcos a, vtsin a -gt> the position vector is <vtcos a, vtsin a -(1/2)gt^2 +h>. R=(v^2 sin 2a)/g t=(2vsin a)g h=(v^2 sin^2 a)/g

  3. anonymous
    • 5 years ago
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    Oh, okay, this looks like projectile motion. Just give me a sec.

  4. anonymous
    • 5 years ago
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    I'll just tell you what I'm doing and you can do it too - you're being asked to find the time of flight. Given your equations, this is going to be the time elapsed between being fired (t=0, y=h) and when it strikes the ground (t=?, y=0).

  5. anonymous
    • 5 years ago
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    k

  6. anonymous
    • 5 years ago
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    Are you sure there's a minus between the 2h and 2Rtan(a)?

  7. anonymous
    • 5 years ago
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    yep

  8. anonymous
    • 5 years ago
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    The problem is, if there is a minus instead of a plus, that expression you have is complex. When you evaluate the radicand, it's negative, so you end up with an imaginary number.

  9. anonymous
    • 5 years ago
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    \[h-R \tan \alpha=\frac{v^2\sin^2 \alpha}{g}-\frac{v^2\sin 2 \alpha}{g}\tan \alpha\]\[=\frac{v^2\sin^2 \alpha -v^2(2\sin \alpha \cos \alpha) }{g}\frac{\sin \alpha}{\cos \alpha}\]\[\frac{-v^2\sin^2 \alpha}{g}\]Multiplying that by 2 and taking the square root will give you\[\sqrt{\frac{2h-2R \tan \alpha }{g}}=\sqrt{\frac{-2v^2 \sin ^2 \alpha}{g^2}}=\frac{\sqrt{2}v \sin \alpha}{g}i\]

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