# Homework Help: Vector and Free Fall Question

1. Sep 11, 2004

### BlackMamba

Hi again,

My first question has to do with vectors again. But it's not hard. I just can't see my mistake.

The problem involves two displacement vectors. A which is 3.00km due south, and B which points due east. No value is given for B. However the resultant vector is 4.49km.

So of course I'm asked to find B and it's direction. So using Pythagorean's theorem of. R^2 = A^2 + B^2, I found B to be 3.34km. Which is correct. However here comes my problem. I know the direction of the resultant is east of south, but when trying to find theta I keep getting a wrong answer. I just don't understand what I did wrong. My answer for theta was 41.9 degrees but when I submit my answer, it is wrong. Where is my mistake?

Secondly, I have a free fall question..

Here is the problem: A diver springs upward with an initial speed of 1.8 m/s from a 5.0m board.

What is the higest point he reaches above the water?

So, I would know how to solve for this if the diver jumped off of the ground. Unfortunately the diver is already 5.0m above the water so my problem is I don't know how to solve this. LOL

I tried using the equation Y = V^2 - Vo^2 / 2a but my answer couldn't possibly be correct as it wouldn't make sense.

A little help and direction for both problems would be greatly appreciated.

2. Sep 11, 2004

There are 2 different angles you could be solving for. Make sure you are finding the right one, and remember that the sum of these two angles will be 90 degrees.

For the second problem, if you solve for the height the diver goes above the board, you would just have to add 5 meters to get his total height above the water.

3. Sep 11, 2004

### Pyrrhus

If $$\vec{B}$$ is pointing East, then it has an angle of 0 degrees, or are they asking you for $$\vec{R}$$ direction?

My english is not that great, but your interpretation of he is 5 meters above the water already... then why don't you add the distance he will go up when he goes in Free Fall.

Info:
$$A = -9.8 m/s^2 (-g)$$
$$V_{o} = 1.8 m/s$$
$$V = 0$$ His final speed becomes 0 when he cannot keep going up and he will be at max height.

Using the Equation
$$V^2 = V_{o}^2 + 2A(Y - Y_{o})$$
$$0 = V_{o}^2 - 2gY$$ We put the origin on the board so initial position is 0
$$Y = \frac{V_{o}^2}{2g}$$

so Y + 5 is the total distance above the water.

Last edited: Sep 11, 2004
4. Sep 11, 2004

### HallsofIvy

In the first problem, you have a right triangle with one leg of length 3 and hypotenuse of length 4.49. Yes, by the Pythagorean theorem the other leg has length (approximately) of 3.34. You can find the angle by using cos(&theta;)= 3/4.49. I get &theta;= 48.1 degrees (east of south), not 41.9.

"So, I would know how to solve for this if the diver jumped off of the ground. Unfortunately the diver is already 5.0m above the water so my problem is I don't know how to solve this. LOL "

Okay, so how about calculating how high the diver went from his initial position (which you say you can find) and then add 5m?

5. Sep 11, 2004

### BlackMamba

The question is asking for the angle of the $$\vec{R}$$ direction. I've tried finding the answer with cos, tan, and sin and have gotten the same answer with all three methods. I'll try again, perhaps I have been just looking at it too long.

Oh my God. I think I need to stop with this homework for tonight, if I couldn't figure out just adding the 5m to my answer. I cannot believe that I didn't think of that.

Thank you both for your help.

6. Sep 11, 2004

### BlackMamba

Thanks HallsofIvy.... I've been working on about 11 different problems today, all day. I still don't have them all finished but it's late and I feel really stupid for not just thinking to add the 5m.

That could also account for why I kept getting the wrong answer for my first problem. I realized that everytime I was using the same numbers but in opposite postions. I think maybe I should just walk away from these problems for awhile, and then come back to it with fresh eyes and maybe then I can spot my own errors.