Recent content by MuchJokes

I hate you both. Though I did draw a picture with it saying that as it falls when I began the question.

The ball hits first. It takes the ball 1.38 seconds to hit the ground. While it takes the flowerpot 2.75 seconds overall.

As I have said, I've already solved for the times. What I don't know how to do is the seconds part. It's asking me for the distance from the ground when the ball passes the flowerpot. Hurrah for large quantaties of formulea, but I have no idea which ones I should use, or what to even do with...

1. A flowerpot is dropped from the balcony of an apartment, 28.5 m above the ground. At a time of 1.00s after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26.0 m above the ground. The initial velocity of the ball is 12.0 m/s [down]. Does the ball...

The bok was wrong. Ironically, the answer book which our teacher has in his possesion had a different answer. It had the RIGHT answer. There went my monday. Thanks for the help.

http://img227.imageshack.us/img227/4299/physicswa9.jpg I know this is against the rules but the book says the answer is 30 degrees S of W Is this just wrong or something? Or does something change from the transition from velocity to displacement?

Here's what it looks like. I used to Cosine law to find the resultant Velocity, and then used the sine law. Unfortunately, my answer for the top angle was 10 degrees. Which apparently is wrong.

A plane, traveling with a velocity relative to the air of 320km/h [28 S of W], passes over Winnipeg. The wind velocity is 72km/h [s]. Determine the displacement of the plane from Winnipeg 2.0 h later. Cosine Law -> c^2 = b^2 + a^2 - 2*b*a*Cos<C Sine Law- (Sin<A)/a = (Sin<B)/b = (Sin<C)/c...