how do you transform this formula to find angle data?

Question

moongazer · Answer

$$Max.\:\:Height = \frac{-(Vi \sin\theta)^2}{2g}$$

anonymous · Answer

oo physics??

moongazer · Answer

yup

anonymous · Answer

$$\theta=\sin^{-1}(\frac{\sqrt{-2gH_{\max}}}{V_i})$$

moongazer · Answer

are you a physics major?
if yes, maybe you can help me ^_^

moongazer · Answer

how did you do it

anonymous · Answer

lol i'm a chem major n today is my last day here

moongazer · Answer

the part that I don't know is the dividing of sine

anonymous · Answer

$$\sin(\theta)=\frac{\sqrt{-2gH_{\max}}}{V_i}$$

anonymous · Answer

to get rid of the sine u take the inverse sine of each side

moongazer · Answer

I am stuck in H2g=-(Vi sin 0)^2   i don't know whats next

anonymous · Answer

$$\sin^{-1}$$
is inverse sine not dividing by sine

anonymous · Answer

multiply each side by -1

anonymous · Answer

then square root it

anonymous · Answer

to get rid of the squared

moongazer · Answer

i am now stuck in   $$(\sqrt{-H2g} )/vi =\sin \theta$$

anonymous · Answer

thats good. ur at the final part

anonymous · Answer

now u have to inverse sin both sides

moongazer · Answer

how?

anonymous · Answer

$$\sin^{-1}(\frac{\sqrt{-2gH_{\max}}}{V_{i}})=\sin^{-1}(\sin(\theta))$$

anonymous · Answer

sin^(-1) and sin cancels each other out

moongazer · Answer

so it will now be:
$$\sin^{-1}(\frac{\sqrt{-2gH_{\max}}}{V_{i}})=\theta$$

anonymous · Answer

exactly

moongazer · Answer

is $$R = v^2 \sin (2\theta)/g$$
the same as
$$R=−(Vi^2\sinθ)/g$$

???

moongazer · Answer

anonymous · Answer

yes. the difference is if u take g to b negative or not

anonymous · Answer

that will change if its negative or not

anonymous · Answer

wait is the original equation correct?

anonymous · Answer

shouldnt it be sin(2theta)

moongazer · Answer

maybe it is a typo

anonymous · Answer

lol u have to check ur formula's properly