Please check this answer for me (Projectile Motion Question)

[an_mui]

Two points A and B are located on the ground a certain distance d apart. Two rocks are launched simultaneously from points A and B with equal speeds but at different angles. Each rock lands at the launch point of the other. Knowing that one of the rocks is launched at an angle \vartheta> 45 degrees with the horizontal, what is the minimum distance between the rocks during the flight? Express your answer in terms of d and \vartheta

Hm.. this is a very hard problem for me so I hope anyone could help me check this over, as there are lots of rooms for mistakes. Thanks in advance.

4 base equations that I used for solving:
(1) x = V_{0}cos\vartheta t
(2)V_{2x} = V_{0}cos\vartheta
(3)y = V_{0}sin\vartheta t + \frac{1}{2}(-g)(t)^2
(4)V_{2y} = V_{0}sin\vartheta + (-g)(t)

D^2 = (x_{2} - x_{1})^2 + (y_{2} - {y_{1})^2

Rock 1:
x_{1} = V_{0}cos \vartheta t
y_{1} = V_{0}sin \vartheta t - \frac{1}{2}gt^2

Rock 2:
x_{2} = d - V_{0}cos \phi t
y_{2} = V_{0}cos \phi t - \frac{1}{2}gt^2

cos \phi = cos(90 - \vartheta) = sin \vartheta
sin \phi = sin (90 - \vartheta) = cos \vartheta

x_{2} - x_{1} = d - V_{0}(cos \vartheta + sin \vartheta)t
y_{2} - y_{1} = V_{0}(cos \vartheta - sin \vartheta)t

Let A = d, B = -V_{0}(cos \vartheta + sin \vartheta) and c = V_{0}(cos \vartheta + sin \vartheta)

D^2 = (A + Bt)^2 + (Ct)^2
= A^2 + 2ABt + B^2t^2 + C^2t^2
= A^2 + 2ABt + (B^2 + C^2)t

Recall quadratic equation = ax^2 + bx + c
Let a = B^2 + C^2, b = 2AB and c = A^2

Minimum = vertex
x = \frac{-b}{2a}
... y_{min} = a(\frac{-b}{2a})^2 + b(\frac{-b}{2a}) + c
= \frac{ab^2}{4a^2} - \frac{b^2}{2a} + c

... d_{min} = \sqrt\frac{4A^2B^2 - 2(4A^2B^2) + A^2}{4(B+C)}
= \sqrt\frac{-A^2B^2 + A^2(B^2 + C^2)}{B^2 + C^2}
= \sqrt\frac{A^2C^2}{B^2 + C^2}

 

[an_mui]

Substitute for A, B and C

d_{min} = \sqrt\frac{d^2V_{0}^2(cos \vartheta + sin \vartheta)^2}{[-V_{0}(cos \vartheta - sin \vartheta)]^2 + [V_{0}(cos \vartheta - sin \vartheta)]^2}

= \frac{dV_{0}(cos \vartheta + sin \vartheta)}\sqrt { V_{0}^2(1- 2cos \vartheta sin \vartheta) + V_{0}^2(1- 2cos \vartheta sin \vartheta)}

 

[an_mui]

sorry for the separate messages.. but the preview keeps showing me my previous equations -_-

= \frac{dV_{0}(cos \vartheta + sin \vartheta)}{V_{0} \sqrt 2 ( 1 - 2 cos <br /> \vartheta sin \vartheta)}

= \frac{d(cos \vartheta + sin \vartheta)}{\sqrt 2 ( 1 - 2 cos <br /> \vartheta sin \vartheta)}

 

[Päällikkö]

With d = 1 m, and \vartheta = 40 degrees, from your equations I get the minimum distance as 66 m, which does seem a bit too big.

In the quadratic equation you've assumed the distance hits zero. This is not the case:
A^2 + 2ABt + (B^2 + C^2)t^2 = D^2 \ne 0

So, your c should be: c = A² - D²

EDIT:
I got |\mathbf{r}|_{min} = \frac{d}{2}\sqrt{(\sin(2 \theta)-1)^2+(2 \sin ^2 \theta - 1)^2}
But I'm not too convinced :)

 

[an_mui]

oh yes 66m does seem unreasonably big.

 

[Päällikkö]

Oh, I suppose there's no use trying to solve t from the quadratic equation (you'd just end up with the same variables).
You should rather figure out a way to compute t when distance is at its minimum.

Hint: There's a mathematical method for this.