- #1
Jacobpm64
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Homework Statement
If a projectile is fired from the origin of the coordinate system with an initial velocity [tex] v_0 [/tex] and in a direction making an angle [tex] \alpha [/tex] with the horizontal, calculate the time required for the projectile to cross a line passing through the origin and making an angle [tex] \beta < \alpha [/tex] with the horizontal.
Homework Equations
[tex] \mathbf{F} = m \mathbf{\ddot{r}} [/tex]
[tex] tan(\theta) = \frac{y}{x} [/tex]
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,
[tex] \ddot{x} = 0 [/tex]
Assuming that [tex] x = y = 0 [/tex] at [tex] t = 0 [/tex] because the projectile is fired from the origin, we get (by integrating)
[tex] \dot{x} = v_{0} cos(\alpha) [/tex]
[tex] x = v_{0} t cos(\alpha) [/tex]
y-direction
We only have one acceleration in the y-direction (gravity).
[tex] \ddot{y} = -g [/tex]
By integrating , we get:
[tex] \dot{y} = -gt + v_{0} sin(\alpha) [/tex]
[tex] y = \frac{-gt^2}{2} + v_{0} t sin(\alpha) [/tex]
Now is where my confusion starts. I suppose I need to solve something of the form [tex] arctan(\frac{y}{x}) < \alpha [/tex] for [tex] t [/tex].
When substituting for [tex] y [/tex] and [tex] x [/tex] and doing some algebra, I get:
[tex] arctan(\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) < \alpha [/tex]
Is there any way to solve this for [tex] t [/tex]? 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.