# 2D Motion

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 50.0 m/s at an angle of 40.0 degrees above the horizontal. What is the range of the object?

2. Relevant equations
All the 2D Motion equations - too many to write all out. Ex:

$$V_x = V_0x + a_xt$$

3. The attempt at a solution
I don't even think there is enough information given to solve this. I can only solve for
$$V_x$$ which is $$50 * cos 40 = 38.3$$ from there I don't know where to go. I try to figure out the y components of this motion, but we don't really know anything about y. We can't really find $$V_y$$ because to find that we need the equation $$V_y = v_0 sin \theta - gt$$ but we don't know the time.

2. Sep 20, 2009

### Kalvarin

To work out the range we need to work out the flight time. So first find the time it takes the ball to reach it's highest point using the y-component of it's velocity. Then calculate how long it takes to hit the floor. The sum of these two times is the flight time.

3. Sep 20, 2009

### Want to learn

ARRRGGGHHHHH - silly me. How could I miss that?

Thank you for pointing me in the right direction, much appreciated m8.

4. Sep 20, 2009

### Kalvarin

Np :)

5. Sep 20, 2009

### RoyalCat

An simpler, more fruitful approach would be to develop two equations, and then turn them into one.

One would be $$x(t)$$, the other would be $$y(t)$$ and combining them will give you $$y(x)$$

That function will describe the height of the object above the the origin as a function of its x-axis distance from the origin. Once you have that function, you can just set $$y=0$$ to find the x for which this holds true (The range.)

To get you started, I'll just rewrite your equation:

$$x(t)=v_0\cos{\theta}\cdot t$$

$$t=\frac{x}{v_0\cos{\theta}}$$

From here, you should be all set.

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