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anonymous
 5 years ago
show that flight time can be estimated by t=sqrt ((2h2Rtan a)/g)
anonymous
 5 years ago
show that flight time can be estimated by t=sqrt ((2h2Rtan a)/g)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's the question that leads to this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, okay, this looks like projectile motion. Just give me a sec.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure there's a minus between the 2h and 2Rtan(a)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[hR \tan \alpha=\frac{v^2\sin^2 \alpha}{g}\frac{v^2\sin 2 \alpha}{g}\tan \alpha\]\[=\frac{v^2\sin^2 \alpha v^2(2\sin \alpha \cos \alpha) }{g}\frac{\sin \alpha}{\cos \alpha}\]\[\frac{v^2\sin^2 \alpha}{g}\]Multiplying that by 2 and taking the square root will give you\[\sqrt{\frac{2h2R \tan \alpha }{g}}=\sqrt{\frac{2v^2 \sin ^2 \alpha}{g^2}}=\frac{\sqrt{2}v \sin \alpha}{g}i\]
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