What Angle Was the Second Projectile Fired to Collide with the First?

[flyingpig]

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?

 

[mathfeel]

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.

 

[bjd40@hotmail.com]

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.

 

[flyingpig]

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?

 

[rwisz]

As a scientist, my response would be that your approach to solving this problem is correct and logical. However, the issue with finding a general solution for t1 is that there are infinite possible values for t1 that could satisfy the equations. In order to find a specific value for t1, you would need more information or equations to solve for it. Alternatively, you could use a numerical method such as trial and error to approximate a solution. Additionally, the angle for the second projectile can also be found using the equations you have set up, but again, a specific value cannot be determined without more information. Overall, your approach is valid and the equations you have set up are correct, but further information or calculations are needed to find a specific solution.