Question

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

Answer 1

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

the same as
\[Max.\:\:Height = \frac{u^2\sin^2\theta}{2g}\]

Answer 2

???

Answer 3

what does i stands for

Answer 4

initial velocity

Answer 5

also the u in 2nd fomula is initial velocity

Answer 6

same

Answer 7

because that is the formula my teacher gave me

Answer 8

it is correct

Answer 9

and same as the one I gave you

Answer 10

I use u for initial velocity and v for final

Answer 11

ok thanks

Answer 12

moongrazer how did your exam go

Answer 13

for university

Answer 14

w8

Answer 15

why is it i am having a different answer?

Answer 16

could you check it?

Answer 17

i am answering the question here http://openstudy.com/groups/physics#/groups/physics/updates/4e4cc3070b8b9588049ae330

Answer 18

I will try

Answer 19

when i use the formula my teacher gave me it is 52.70
and if i tried yours it is 43.63

Answer 20

I am getting 51.7

Answer 21

try it again must be missing something it is square

Answer 22

on both?

Answer 23

no max height I am talking about max height let me do range too

Answer 24

no, i mean do you get 51.7 on using both formulas?

Answer 25

246.20 range
51.7 max height

Answer 26

did you used the formula my teacher gave me?

Answer 27

it's the same answer is same 51.7 and 246.20
I am using calculator the angle is 40 and initial velocity is 50

Answer 28

^^^^^^^^^even with your teacher's formula

Answer 29

ohhh i had the correct answer now

Answer 30

good job : )

Answer 31

i squared 50 that is why i got a wrong answer

Answer 32

Thanks for helping me :)

Answer 33

anytime : )

Answer 34

physics have a lot of formulas

Answer 35

We have an index card that contains i think more or less 10 formulas that we will use for our exam. But i think we will not gonna use it all. ^_^

Answer 36

Good Luck ...just practice questions and then you won't have problem remembering formulas

Answer 37

yup, practices makes perfect

Answer 38

i have a last question

Answer 39

okay I will try that in 5 minutes I'm hungry now gotta eat somthing

Answer 40

how do you transform an equation if you want to have the formula for\[\theta\]
like \[R= (-Vi^2 \sin \theta)/g\]

please explain how
thanks for helping me :)
\[Vi = initial velocity\]

Answer 41

is it the same as\[R = v^2 \sin (2\theta)/g\]

Answer 42

http://openstudy.com/groups/physics/updates/4e4cc49d0b8b9588049ae846

Answer 43

??

Answer 44

Okay I am back.

\[R = \frac{V_i^2 sin2\theta}{g}\]
\[R \times g = V_i^2 sin2\theta\]
\[\frac{R \times g }{V_i^2}=sin2 \theta\]
\[sin^{-1} \frac{R \times g }{V_i^2} = 2\theta\]

\[\frac{1}{2}sin^{-1} \frac{R \times g }{V_i^2} = \theta\]