(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A ball is kicked off of a hill with an initial horizontal velocity of 10 m/sec.

The pitch of the hill from where the ball is thrown is -20°. Determine how far the ball lands from its original position.

Not given, but I'm acting under the assumption that gravity = -9.81m/sec²

This is how I see the problem,

http://img78.imageshack.us/img78/6411/projectileeu0.gif [Broken]

2. Relevant equations

a = v/t

d = vo*t + ½at²

3. The attempt at a solution

I would love to say that;

a = v/t

9.81 = 10/t

t = 1.02sec

Then just split and solve so that

d(x) = 10.2, d(y) = -5.1

√(dx² + dy²) = 11.4m

But I can't do that because there is no y velocity so, solving for the only constant.

d = vo*t + ½at²

Split into x,y.

d(y) = -4.905 t²

t = √(d(y)/- 4.905)

d(x) = vo(x)*t

t = d(x)/10

Setting them equal,

d(x)/10 = √(d(y)/- 4.905)

dx = 10√(dy/- 4.905)

dy = - 4.905(dx/10)²

To find thedistance from startuse law cosines, or in this case pythag.

d = √[{10√(dy/- 4.905)}² + {- 4.905(dx/10)²}²]

While this is nice, it isn't getting me any closer to the answer. I need to solve for time somehow...

So... If anyone can give me a nudge it would be extremely grateful.

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# Projectile Motion Off a Slope.

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