# Homework Help: Finding when a projectile has a certain angle with the horizontal

1. Aug 27, 2009

### Jacobpm64

1. The problem statement, all variables and given/known data
If a projectile is fired from the origin of the coordinate system with an initial velocity $$v_0$$ and in a direction making an angle $$\alpha$$ with the horizontal, calculate the time required for the projectile to cross a line passing through the origin and making an angle $$\beta < \alpha$$ with the horizontal.

2. Relevant equations
$$\mathbf{F} = m \mathbf{\ddot{r}}$$

$$tan(\theta) = \frac{y}{x}$$

3. The attempt at a solution
So, here I go. First, I treated the x-direction and y-direction separately.

x-direction
I know that there is no acceleration in the x-direction so,

$$\ddot{x} = 0$$

Assuming that $$x = y = 0$$ at $$t = 0$$ because the projectile is fired from the origin, we get (by integrating)
$$\dot{x} = v_{0} cos(\alpha)$$

$$x = v_{0} t cos(\alpha)$$

y-direction
We only have one acceleration in the y-direction (gravity).

$$\ddot{y} = -g$$

By integrating , we get:

$$\dot{y} = -gt + v_{0} sin(\alpha)$$

$$y = \frac{-gt^2}{2} + v_{0} t sin(\alpha)$$

Now is where my confusion starts. I suppose I need to solve something of the form $$arctan(\frac{y}{x}) < \alpha$$ for $$t$$.

When substituting for $$y$$ and $$x$$ and doing some algebra, I get:

$$arctan(\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) < \alpha$$

Is there any way to solve this for $$t$$? Or, is there another way of working this problem that I am not seeing?

It's been quite a long time since I've worked with mechanics, and this is my first upper-level mechanics course, so I don't really know what I'm dealing with here.

Any help is appreciated. Thanks in advance.

2. Aug 27, 2009

### kuruman

I have not checked your algebra, but if

$$arctan(\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) < \alpha$$

is it not true that

$$(\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) < tan\alpha ?$$

Can you now solve for the time?

3. Aug 27, 2009

### Jacobpm64

Wow, It's amazing how a senior math major can overlook simple things when he's afraid.

So, that gives $$t > 0$$? I guess that makes sense actually. I kinda sketched the graphs of a bunch of projectile motions, and as long as they're concave down (which should always be the case assuming gravity) then the largest angle with respect to the horizontal is infact $$\alpha$$ at $$t = 0$$, and any angle when $$t > 0$$ is less than $$\alpha$$. Does this sound like a good answer?

"When $$t > 0$$."

4. Aug 27, 2009

### kaymant

It's probably much easier if you take your $$+x$$ axis along the line through the origin inclined to the horizontal at the given angle $$\beta$$ and the $$+y$$ axis perpendicular to it, like in the figure.

Then, we have the acceleration components:
$$a_x=-g\sin\beta,\quad a_y=-g\cos\beta$$
and the initial velocity components:
$$v_{0x}=v_0\cos(\alpha-\beta),\quad v_{0y}=v_0\sin(\alpha-\beta)$$
We require the time interval between the points where it crosses the x-axis. Obviously, it will be the same time during which the y coordinate returns to zero. Now the y coordinate of the particle
$$y=v_{0y}t+\dfrac{1}{2}a_yt^2$$
So $$y=0$$ imply
$$v_{0y}t+\dfrac{1}{2}a_yt^2=0$$
the roots of which are
$$t_1=0,\quad t_2=-\dfrac{2v_{0y}}{a_y}=-\dfrac{2v_0\sin(\alpha-\beta)}{-g\cos\beta}=\dfrac{2v_0\sin(\alpha-\beta)}{g\cos\beta}$$
The required time is same as t2.
Hence, the answer: $$\boxed{\dfrac{2v_0\sin(\alpha-\beta)}{g\cos\beta}}$$

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5. Aug 27, 2009

### Jacobpm64

It's interesting how simple a problem becomes when you just change the coordinate system.

Thanks a lot!