# Vectors and 2 dimensional motion, driving me nuts

1. May 29, 2007

### skjhlkj

1. The problem statement, all variables and given/known data
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 21.0 m/s. The cliff is h = 46.0 m above a flat horizontal beach
Initial v =21.0m/s
Delta Y= 46.0
Need to know, How long after being released does the stone strike the beach below the cliff? & With what speed and angle of impact does the stone land?

2. Relevant equations
I calculated the time = 3.06 which I got it right even though I didn't understand it. Square root of (46/4.9)=3.06s
The formula I used was "Delta Y=Vyo*T - (1/2)*g*(t square).
For some reason Initial velocity of Y was counted zero, eventhough it was given (21m/s).

The final velocity is 36.6. I don't understand How i get it right.
Vy = 21 + (-9.8)*5.88 => Vy = 36.6 m/s, Eventhought 5.88s was not the accurate answer for time (3.06s should be), but somehow 36.6 m/s was the right answer.

3. The attempt at a solution

I'm really having trouble finding the angle of impact below the horizontal.
First, I don't know what "below the horizontal means.
Second, I don't know how to calculate degree from inverse sine. I got to
.3149=sine of theta. The answer is close the 60degree, but I couldn't get it right.

Man, phsyics is tough. It's not just the math, but my textbook doesn't really explain much. Maybe I'm a bit slow, but the textbook doesn't show every step. It skip steps, jump around formulas and doesn't show all the work. Is there any website out there that shows detail steps? (for intro physics) :yuck:
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 29, 2007

### husky88

There are many websites. Look in the forums under the tutorials. Also you can just google for a topic you don't understand.

"and throws a stone horizontally over the edge with a speed of 21.0 m/s"
Horizontal speed is 21. It says horizontally. What is the speed vertically then?

In two dimensions you have to find the two components of the velocity. There is vx horizontally and vy vertically.

Last edited: May 29, 2007
3. May 29, 2007

### andrevdh

The kinematics of objects following parabolic trajectories can be described by considering the horizontal and vertical motion separately. The horizontal motion progresses undisturbed by the gravitational attraction by the earth. The horizontal speed of the object remains constant. The vertical motion of the object is described by the constant acceleration equations. As time progresses one can calculate the object's x and y coordinates with these two separate motions.