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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

What's the question that leads to this?
There isn't one but the information that i have is that the velocity vector is the position vector is . R=(v^2 sin 2a)/g t=(2vsin a)g h=(v^2 sin^2 a)/g
Oh, okay, this looks like projectile motion. Just give me a sec.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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).
k
Are you sure there's a minus between the 2h and 2Rtan(a)?
yep
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.
\[h-R \tan \alpha=\frac{v^2\sin^2 \alpha}{g}-\frac{v^2\sin 2 \alpha}{g}\tan \alpha\]\[=\frac{v^2\sin^2 \alpha -v^2(2\sin \alpha \cos \alpha) }{g}\frac{\sin \alpha}{\cos \alpha}\]\[\frac{-v^2\sin^2 \alpha}{g}\]Multiplying that by 2 and taking the square root will give you\[\sqrt{\frac{2h-2R \tan \alpha }{g}}=\sqrt{\frac{-2v^2 \sin ^2 \alpha}{g^2}}=\frac{\sqrt{2}v \sin \alpha}{g}i\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question