# Homework Help: Dynamics in two dimensions problem

1. Oct 15, 2016

### BrainMan

1. The problem statement, all variables and given/known data

2. Relevant equations
xf = x0 + v0cos(θ)t + (1/2)axt2
yf = y0 + v0sinθt - (1/2)gt2
F = ma
3. The attempt at a solution

I first tried to find the acceleration in the x direction so I could solve for the force
xf = x0 + v0cos(θ)t + (1/2)axt2

In order to find the time I used the equation for the y coordinate
yf = y0 + v0sinθt - (1/2)gt2

The initial and final y positions will both be zero so
0 = 0 + v0sin(θ) - (1/2)gt2
t = 2v0sin(θ)/ g = .702 sec

Then I plugged time back into the x equation

2.9 = v0cos(θ)(.702) + (1/2)a(.702)2
a = -2.23

Since I know the acceleration I can find the force by Newtons second law

F = ma

F = .015(-2.23) = -.033 N

It appears that I am a sign off from the correct answer

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2. Oct 15, 2016

### J Hann

0 = 0 + v0sin(θ) - (1/2)gt2 t = 2v0sin(θ)/ g = .702 sec

Don't you need to take the square root here?

3. Oct 15, 2016

### PeroK

What do you mean by "a sign off"?

4. Oct 15, 2016

### BrainMan

The correct answer is 0.034 N so it looks like I'm getting the negative of the right answer.

5. Oct 15, 2016

### PeroK

And what does the question ask for?

6. Oct 15, 2016

### BrainMan

Oh I see. Since it's asking for the magnitude it's just asking for the value and not the sign, right?

7. Oct 15, 2016

### PeroK

Well, I would say it's asking for the magnitude! Which is always a positive number. The negative sign is actually a direction: opposite the direction of motion, in this case.

"Value" suggests magnitude and direction to me.