What is the missing formula for projectiles launched at an angle?

[xxpbdudexx]

I looked at the sticky, but could not find what I needed. I got a trio of new formulas today, but I never copied them down. I remembered the first two because I used them today, but I cannot recall the other one. The subject is Projectiles Launched at an Angle and the 2 I have are:

R= (vi²+2sin$$\theta$$)/g
y_max=(vi²+sin²$$\theta$$)/2g

 

[rl.bhat]

Both are wrong.
Instead of memorizing the formula, try to understand how they are derived.
Go to Hyper Physics site and get all the information.

 

[xxpbdudexx]

Both are wrong.
Instead of memorizing the formula, try to understand how they are derived.
Go to Hyper Physics site and get all the information.

What do you mean both are wrong?

And also what do you mean by understanding the derivation? Will that help me in some way?

I just need to know this missing formula that I didn't copy down.

 

[rl.bhat]

What are basic thing you know about the projectile motion?
Have you gone through any textbook?
Why are you talking about missing formula?
You have written the formula for range and maximum height. Which remaining quantity you have to know about the projectile motion?

 

[xxpbdudexx]

What are basic thing you know about the projectile motion?
Have you gone through any textbook?
Why are you talking about missing formula?
You have written the formula for range and maximum height. Which remaining quantity you have to know about the projectile motion?

1. Um, at the highest point of the arc, the v=0, and there are 2 times for 1 height.
2. I don't have one.
3. My teacher gave us 3 formulas and I didn't copy them down. I need the other formula I can't remember.
4. I'm not sure. Sorry I have no more information.

 

[rl.bhat]

Range R = v^2*sin(2θ)/g
Maximum height =ymax = v^2sin^2(θ)/2g
Time of flight T = 2vsin(θ)/g

 

[xxpbdudexx]

Range R = v^2*sin(2θ)/g
Maximum height =ymax = v^2sin^2(θ)/2g
Time of flight T = 2vsin(θ)/g

Third one is it. Thanks.