# Homework Help: 2D Motion part deux

1. Sep 20, 2009

### Want to learn

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

An object is shot from the origin with a velocity of 40.0 m/s at an angle of 55.0 degrees above the horizontal. What is the location of the object 3.00 seconds later?

2. Relevant equations

All 2D motion equations - too many to list Ex:

$$V_x = v_0x + a_x t$$

$$V_y = v_0y + a_y t$$

3. The attempt at a solution
What I am having trouble with is the 3.0 seconds later part. I am not sure how to approach this one. I have to find the x displacement and then the angle $$\theta$$ which is no problem to do when I know $$V_y and V_x$$.

So I tried this:

$$\Delta x = 40 cos(55)3.0 = 68.8m$$ , but none of the answer choices match this. So I must be doing something wrong. I also can't get a correct answer for $$V_y and V_x$$ with these I could plug them into the inverse tangent equation and get an angle. To find $$V_x$$ I do this:

$$V_x = 40 cos (55) = 22.9$$

for $$V_y$$ I do this :

$$V_y = 40 sin (55) - (9.80)(3.0^2) = -55.4$$

I am not sure that I am using the time variable correctly.

2. Sep 20, 2009

### sylas

You can solve the horizontal and vertical parts independently.

Your Δx of 68.8 looks fine.

Your final equation for Vy is wrong. Are you calculating a velocity, or a distance?

Under relevant equations, you've given a correct formula for getting a velocity, given initial velocity and acceleration.... but it doesn't have a t2 factor anywhere. The question, of course, requires a distance.

Do you know an equation for distance, given an initial velocity, an acceleration, and a time?

Cheers -- sylas

3. Sep 20, 2009

### Want to learn

well there is this one:

$$\Delta x = V_0_x t + (1/2) (a_x) t^2$$

but $$a$$ in the horizontal direction = 0 in this case. So I am back to where I started.

I am trying to find, where the object is going to end up after 3.00 seconds.

I need the distance traveled in 3 seconds. And then I need to find the angle that vector makes with the x-axis using the inverse tangent equation. But to use the tangent equation I need to find $$V_y and V_x$$

These are the equations from my book:

$$V_y = V_0 sin \theta - gt$$
$$V_x = V_0 cos \theta$$

when I use these, it doesn't work out.

4. Sep 20, 2009

### Redbelly98

Staff Emeritus
That correctly gave you x=68.8 m. If you use the analogous equation for Δy, you'll get y as well.

Since the question asks for the location of the object, all you need is x and y. We don't need Vx, Vy, or the distance travelled.