Maybe it's because I'm over-analyzing it but I've wasted 20 sheets of paper trying to get the right answer for this problem, but I just dont get it!!!!!

A shot put is released with a velocity of 12 m/s and stays in the air for 2.0s.

At what angle with the horizontal was it released?

What horizontal distance did it travel?

Question

Maybe it's because I'm over-analyzing it but I've wasted 20 sheets of paper trying to get the right answer for this problem, but I just dont get it!!!!!

A shot put is released with a velocity of 12 m/s and stays in the air for 2.0s.


At what angle with the horizontal was it released?


What horizontal distance did it travel?

anonymous · Answer

$$x=\left( v _{0}\cos \alpha \right)t$$
Where x is the horizontal distance, v0 is the initial velocity and alpha is the angle.

anonymous · Answer

You simply do not have enough info to solve this. 
4 variables and 2 is given. 
The other 2 that you are looking for are dependent of eachother

anonymous · Answer

Wait a sec, I know now!

anonymous · Answer

@Miranda_Esquivel210 Are you here?

anonymous · Answer

yea @Andras

anonymous · Answer

$$y=\left( v _{0}\sin \alpha \right)t -g t^2$$

anonymous · Answer

That is the y coordinate, sorry that took so long. 
From this equation we know everything apart from alpha!

anonymous · Answer

Solve that and you get alpha. Once you have it solve the first one and you have x

anonymous · Answer

omg i've been trying to work this for a week you're amazing! @Andras

primeralph · Answer

20 sheets? That's impressive.

anonymous · Answer

At least you have it now. I hope you do not me to derive the above equations

anonymous · Answer

*do not need me

anonymous · Answer

no i'm fine, i got it now

anonymous · Answer

@primeralph most of it is basic physics, this question just through me off

primeralph · Answer

Okay.
*threw.