# Educate me please, I have to find initial velocity!

1. Sep 18, 2009

### exparrot

I'm doing a physics lab write-up. It was a projectile lab launching the projectile horizontally. The projectile was launched at a higher elevation (the table) down to the ground.

1. The problem statement, all variables and given/known data

Calculate how fast the projectile came out of the launcher. Use the data obtained.

Well, my only data is the distance traveled by the projectile which is 3.7870 m.

2. Relevant equations

R = √(Rg)/(sin 2θ) ---> since measuring time was not part of the lab, this is the only other equation with no extra unknown variables

3. The attempt at a solution

Don't know how!

How can I solve this given that I only know my distance? I don't know my initial velocity (which I have to find), don't know my final velocity (hesitant to say it's 0 m/s) and my acceleration? I don't think the gravitational acceleration is information integral to this situation since it's horizontally launched. Would supremely appreciate any help, thanks!

2. Sep 18, 2009

### sylas

How high was the table? If you didn't measure this, you can't get the result. Just knowing the distance, as you observe, is not going to be enough.

Cheers -- sylas

3. Sep 18, 2009

### exparrot

0.775 m

4. Sep 18, 2009

### drizzle

what do R and √(Rg) stand for?

ps. the final velocity of a projectile will never be zero, drill that into your mind

5. Sep 18, 2009

### exparrot

R is the range/distance and g is acceleration due to gravity. The original equation was this:

R = [(v0^2)(sin 2θ)]/g

where v0 is the initial velocity. I can't do subscripts here, not sure if I can...

6. Sep 18, 2009

### sylas

That now gives you enough to solve the problem. The next step is part 2: the relevant equations you are using. If you use variables, it will help to say what they represent; as drizzle has also noted.

There may be other equations you should consider...

PS. OK. That's better, but I think you may be using an equation for the range on a flat surface. I've checked; yes, this is not the equation you really want here.

To get subscripts and super scripts, use sub and sup tags.

For example.

input: [noparse]R = (v02 sin 2θ) / g[/noparse]
result: R = (v02 sin 2θ) / g

Last edited: Sep 18, 2009
7. Sep 18, 2009

### exparrot

I can use the kinematics equations, but my only problem is A) don't know the time as our prof didn't ask us to do that, B) do not know the initial or final velocity. That leaves me with alot of unknown variables to work with. The speed I calculate in this question is suppose to help me solve subsequent questions.

8. Sep 18, 2009

### sylas

Here's a clue. If you launch something horizontally, it takes the same amount of time to reach the ground as if you just dropped it, because the acceleration downwards is still the same, and the initial velocity downwards is 0 in both cases.

9. Sep 18, 2009

### drizzle

edit: ops not for this one :)