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So, is anyone out there good at physics? My teacher gave us a story problem for a ton of extra credit, and I'd really like to get it right. I need to be able to explain how I got the answer too though.
Here's the problem:
"A soccer ball is kicked on a level playing field with an initial velocity of 12 m/s and stays in the air for 2.0 s before hitting the ground."
I have to find the distance it traveled horizontally, and what angle it was initially kicked at.
I do know that the acceleration due to gravity will be -10 m/s2, (because we're on earth) and that the final velocity will be -12 m/s because it will come down at the same speed in the opposite direction. Other than that, the problem gives no other information…

I'm normally pretty good at physics, but we aren't given the angle of trajectory in this problem. I've looked at it every way I can, but I still can't figure it out.
Help me? TT_TT

This might be hard to describe over the internet, but what the hell, I'll give it a shot.
Keep in mind that using this answer is totally cheating. If you've got problems with that, you might want to stop reading now.

First, consider only the vertical component of motion. Since you know it's only airborne for 2 seconds, and that acceleration due to gravity is -10 m/s^2, you can use the equation below to work out the ball's initial vertical velocity:

s = u t + (1/2) a t^2
where:
s = final displacement = 0
t = time elapsed = 2
a = acceleration due to gravity = -10

Stick those numbers into the equation and you should find that the initial vertical velocity is 10m/s.

From here, you can work out the angle of trajectory

You know that A=12 and Ay=10, so using the equation sinθ=Ay/A you can work out that θ=56.443

Now you can work out the horizontal velocity, by Ax=Acosθ. You should find that Ax=6.63m/s.

To get to the final answer, simply use the equation distance=speed x time
distance = 2 × 6.63
distance = 13.26 m

Damn it. I knew it would be something stupid I over looked. __
Anyways, thanks a ton! I just needed to know how to get one of the velocities for x or y, then I could do the rest on my own.