# Kinematics, with two unknowns

## 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.

$$y_1 (t) = \frac{-gt_1^2}{2} + 30\sqrt{3}t_1$$

$$y_2 (t) = \frac{-g(t_1 + \frac{1}{2})^2}{2} + 60(t_1 + \frac{1}{2})\sin\alpha$$

$$x_1 (t) = 30t_1$$
$$x_2 (t) = 60(t_1 + \frac{1}{2})\cos\alpha$$

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

$$y_1 (t) = y_2 (t)$$

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

(2) $$(30\sqrt{3} + \frac{g}{2})t_1 + \frac{g}{8} = 60(t_1 + \frac{1}{2})\sin\alpha$$

$$x_1 (t) = x_2 (t)$$

(1) $$30t_1 = 60(t_1 + \frac{1}{2})\cos\alpha$$

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

$$\frac{(30\sqrt{3} + \frac{g}{2})t_1 + \frac{g}{8}}{30t_1} = \tan\alpha$$

$$\sqrt{3} + \frac{g}{60} + \frac{g}{240t_1} = \tan\alpha$$

Now here is the problem, what is t1? Will all values (other than 0) of t1 work?

I tried to go to back to my other basic kinematics equation, but I don't want a particular t1, I need a general one?

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Now here is the problem, what is t1? Will all values (other than 0) of t1 work?

I tried to go to back to my other basic kinematics equation, but I don't want a particular t1, I need a general one?
You have two equations and two unknown. So it can be solved. Just solve $t$ in term of $\alpha$ from one equation and substitute it into the second equations.

BTW, you original equations are wrong. the trajectory for $x_2$ and $y_2$ should have $t_1 - 0.5$ instead of $t_1+0.5$. The travel time for the second projectile is less than the first one.

first, i think u have put it wrong, the time of 2nd projectile, it has been fired 1/2 sec later, so it will take less time to collide or travel the same distance with the first. so the time will be t1-1/2. 2nd-ly u can always find t1 in terms of cos alfa from the eqn. x1(t1) = x2(t2). replace t1 in terms of cos alfa in the eqn. u got by deviding eqn. 1 with 2. there will be only one unknown, alfa. it can be solved with a bit of mathematics. i have not worked it out myself, so i don't know how cumbersome it may be.

So I got

$$\sqrt{3} + \frac{g}{240t_1} - \frac{g}{60} = \tan\alpha$$

From x_1 = x_2

$$t_1 = \cos\alpha$$

$$\sqrt{3} + \frac{g}{240\cos\alpha} - \frac{g}{60} = \tan\alpha$$

Any know any trig identities to simplify this?