Question

How would you solve for theta:
R= 3h
when R= (v^2*sin2theta)/g
and h=(v^2*sin^2theta)/g

Answer 1

so far I have
(v^2*sin2theta)/g = (3v^2*sin^2theta)/2g

Answer 2

I then put a common denominator of 2g and finally in the end where I find myself stuck is

2sin2theta=3sin^2theta

don't know if perhaps I did something wrong and I just don't know what to do now

Answer 3

Is this a typo, or is h=R? I'm basing this off of your definition in the OP.

Answer 4

now that I say that though, would it instead be 3R=h? since R is 3X bigger

Answer 5

In that case, what you want to do is to solve for \(\theta\) in the equation\[\frac{3\sin^2(\theta)v^2}{2g}=\frac{v^2\sin(2\theta)}{g}.\]

Answer 6

what just happend :o

Answer 7

but yes that is what I'm trying to figure out

Answer 8

\[3\sin^2(\theta)=g\sin(2\theta)\]\[\frac{\sin^2(\theta)}{\sin(2\theta)}=\frac{g}{3}\]\[\frac{\sin^2(\theta)}{2\sin(\theta)\cos(\theta)}=\frac{g}{3}\]\[\tan(\theta)=\frac{2}{3}g\]

Answer 9

I'm curious as to how you got the first part. The 3sin^2=gsin(2theta)

Answer 10

I just multiplied both sides by \(2g\).

Answer 11

Then divided both sides by \(v^2\).

Answer 12

wouldn't the g's "cancel out? As in the value of g would no longer be in the formula after that

Answer 13

and is it an identity or something? sin2/2sin*cos = tan?

Answer 14

The \(g\)'s do cancel out: The LHS cancels out completely, whilst the RHS only partially, i.e.,\[\frac{2g}{g}=g.\]I'm also using the identities\[\sin2u=2\sin u\cos u\]and\[\frac{\sin u}{\cos u}=\tan u\]

Answer 15

ok, I do see how you got the tan; as for the LHS and RHS; sorry, but I'm lost theree. How is it canceled partially?

Answer 16

\[\frac{3\sin^2(\theta)v^2}{2g}=\frac{v^2\sin(2\theta)}{g}\]\[\frac{2g}{2g}3\sin^2(\theta)v^2=\frac{2g}{g}v^2\sin(2\theta)\]\[3\sin^2(\theta)v^2=gv^2\sin(2\theta)\]\[3\sin^2(\theta)=g\sin(2\theta)\]

Answer 17

LOL I realized I made a mistake.

Answer 18

I'm sorry for confusing you.

Answer 19

You're right, it should be \(2\), not \(g\)!

Answer 20

lol, it's ok, I was like how is he getting that :o

Answer 21

still thanks for everything; you explained the problem out which helped a lot!! :D

Answer 22

Haha, I'm sorry. :3 That was SO silly of me. I will now go and hide my face in shame. LOL

Answer 23

nah, don't worry, I a few days back I added 170 and 44 and obtained 116. Who should hide there face more? LOL

Answer 24

LOL I guess we all have our off days then. :)

Answer 25

Thanks again! :)

Answer 26

yw :)