# Find the speed v0 and height

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1. Oct 30, 2016

### annalian

1. The problem statement, all variables and given/known data
The particle thrown under the angle pi/6 rad. with the horizontal direction achieves the height h for t1=10S when going up and t2=50S when going down. Find initial speed v0 and height h.

2. Relevant equations
h=v0y+gt^2/2

3. The attempt at a solution
As h1=h2
v0y*t1-gt1^2/2=gt2^2/2
v*0.5*10-10*100/2=10*2500/2
v0=2600.
hmax:
v^2-vy0^2=2gh
vy0=v0*sin30=2600*0.5=1300
h=0-1300^2=2*(-10)*h
h=84500m.
Am i right?

2. Oct 31, 2016

### Ibix

Where does this expression come from?
Where does this expression come from?

Just below the reply box is a link to a LaTeX guide which you may find helpful for writing formulae. For example $h=v_{0y}t+\frac {1}{2}at^2$.

3. Oct 31, 2016

### annalian

Can you help me where to find the correct expressions for this exercise?

4. Oct 31, 2016

### Ibix

The expression you quoted in part 2 of your question seems sensible, and your reasoning that $h_1=h_2$ seems reasonable. However, I think you need to look at how you got from the equation in part 2 to the first expression I quoted. Post it here if you don't spot anything wrong.

You haven't provided any justification for the second expression at all. Where did it come from? It isn't immediately obvious that it's one of the kinematic formulae. I'm not saying it's wrong, just that I don't know where you got it from. I can guess, but it's easier if you tell me.

5. Oct 31, 2016

### annalian

v^2-vy0^2=2gh
As we know v^2-v0^2=2gh, but as we have two vectors of vo we should only use the one voy.

6. Oct 31, 2016

### Ibix

Do we know that? Why? Neither of the expressions you've written makes sense to me, and you aren't explaining why you think they should.

Start with a statement of the law or principle or whatever you think you are applying to get those equations.