Finding when a projectile has a certain angle with the horizontal

[Jacobpm64]

Homework Statement

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.

Homework Equations

\mathbf{F} = m \mathbf{\ddot{r}}

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

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.

 

[kuruman]

I have not checked your algebra, but if

<br /> arctan(\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) &lt; \alpha <br />

is it not true that

<br /> (\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) &lt; tan\alpha ?<br />

Can you now solve for the time?

 

[Jacobpm64]

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

So, that gives t &gt; 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 &gt; 0 is less than \alpha. Does this sound like a good answer?

"When t &gt; 0."

 

[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}}

 

[Jacobpm64]

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

Thanks a lot!