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## Homework Statement

A projectile is fired at an angle 60 degrees above the horizontal with a speed of 60m/s. A 0.5 second later, another projectile fired at the same speed but at a different angle. The two projectiles collide at some point (x,y). Find the angle the second projectile was fired and the coordinate (x,y) they collide.

## The Attempt at a Solution

All subscripts "1" refer to the first projectile and the "2" refer to the second projectile.

[tex]y_1 (t) = \frac{-gt_1^2}{2} + 30\sqrt{3}t_1[/tex]

[tex]y_2 (t) = \frac{-g(t_1 + \frac{1}{2})^2}{2} + 60(t_1 + \frac{1}{2})\sin\alpha[/tex]

[tex]x_1 (t) = 30t_1[/tex]

[tex]x_2 (t) = 60(t_1 + \frac{1}{2})\cos\alpha[/tex]

Naturally I set them equal to each other to find that (x,y) coordinate

[tex]y_1 (t) = y_2 (t)[/tex]

[tex]30\sqrt{3} t_1 = \frac{-g}{2}(t_1 + \frac{1}{4}) + 60(t_1 + \frac{1}{2})\sin\alpha[/tex]

(2) [tex](30\sqrt{3} + \frac{g}{2})t_1 + \frac{g}{8} = 60(t_1 + \frac{1}{2})\sin\alpha[/tex]

[tex]x_1 (t) = x_2 (t)[/tex]

(1) [tex]30t_1 = 60(t_1 + \frac{1}{2})\cos\alpha[/tex]

So from Euclid's elements, I can now divide (2) by (1)

[tex]\frac{(30\sqrt{3} + \frac{g}{2})t_1 + \frac{g}{8}}{30t_1} = \tan\alpha[/tex]

[tex]\sqrt{3} + \frac{g}{60} + \frac{g}{240t_1} = \tan\alpha[/tex]

Now here is the problem, what is t

_{1}? Will all values (other than 0) of t

_{1}work?

I tried to go to back to my other basic kinematics equation, but I don't want a particular t

_{1}, I need a general one?

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