Dynamics in two dimensions problem

[BrainMan]

Homework Statement

Homework Equations

x_f = x₀ + v₀cos(θ)t + (1/2)a_xt²
y_f = y₀ + v₀sinθt - (1/2)gt²
F = ma

The Attempt at a Solution

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I first tried to find the acceleration in the x direction so I could solve for the force
x_f = x₀ + v₀cos(θ)t + (1/2)a_xt²

In order to find the time I used the equation for the y coordinate
y_f = 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

 

[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?

 

[PeroK]

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

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

What do you mean by "a sign off"?

 

[BrainMan]

What do you mean by "a sign off"?

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

 

[PeroK]

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

And what does the question ask for?

 

[BrainMan]

And what does the question ask for?

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

 

[PeroK]

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

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.