- #1
an_mui
- 47
- 0
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 [tex]\vartheta [/tex]> 45 degrees with the horizontal, what is the minimum distance between the rocks during the flight? Express your answer in terms of d and [tex]\vartheta[/tex]
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:
[tex](1) x = V_{0}cos\vartheta t[/tex]
[tex](2)V_{2x} = V_{0}cos\vartheta[/tex]
[tex](3)y = V_{0}sin\vartheta t + \frac{1}{2}(-g)(t)^2[/tex]
[tex](4)V_{2y} = V_{0}sin\vartheta + (-g)(t)[/tex]
[tex]D^2 = (x_{2} - x_{1})^2 + (y_{2} - {y_{1})^2[/tex]
Rock 1:
[tex]x_{1} = V_{0}cos \vartheta t[/tex]
[tex]y_{1} = V_{0}sin \vartheta t - \frac{1}{2}gt^2[/tex]
Rock 2:
[tex]x_{2} = d - V_{0}cos \phi t[/tex]
[tex]y_{2} = V_{0}cos \phi t - \frac{1}{2}gt^2[/tex]
[tex] cos \phi = cos(90 - \vartheta) = sin \vartheta[/tex]
[tex] sin \phi = sin (90 - \vartheta) = cos \vartheta[/tex]
[tex]x_{2} - x_{1} = d - V_{0}(cos \vartheta + sin \vartheta)t[/tex]
[tex]y_{2} - y_{1} = V_{0}(cos \vartheta - sin \vartheta)t[/tex]
[tex]Let A = d, B = -V_{0}(cos \vartheta + sin \vartheta) and c = V_{0}(cos \vartheta + sin \vartheta)[/tex]
[tex]D^2 = (A + Bt)^2 + (Ct)^2[/tex]
[tex]= A^2 + 2ABt + B^2t^2 + C^2t^2[/tex]
[tex]= A^2 + 2ABt + (B^2 + C^2)t[/tex]
Recall quadratic equation = ax^2 + bx + c
[tex]Let a = B^2 + C^2, b = 2AB and c = A^2[/tex]
Minimum = vertex
[tex]x = \frac{-b}{2a}[/tex]
[tex]... y_{min} = a(\frac{-b}{2a})^2 + b(\frac{-b}{2a}) + c[/tex]
[tex] = \frac{ab^2}{4a^2} - \frac{b^2}{2a} + c[/tex]
[tex]... d_{min} = \sqrt\frac{4A^2B^2 - 2(4A^2B^2) + A^2}{4(B+C)}[/tex]
[tex] = \sqrt\frac{-A^2B^2 + A^2(B^2 + C^2)}{B^2 + C^2}[/tex]
[tex] = \sqrt\frac{A^2C^2}{B^2 + C^2}[/tex]
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:
[tex](1) x = V_{0}cos\vartheta t[/tex]
[tex](2)V_{2x} = V_{0}cos\vartheta[/tex]
[tex](3)y = V_{0}sin\vartheta t + \frac{1}{2}(-g)(t)^2[/tex]
[tex](4)V_{2y} = V_{0}sin\vartheta + (-g)(t)[/tex]
[tex]D^2 = (x_{2} - x_{1})^2 + (y_{2} - {y_{1})^2[/tex]
Rock 1:
[tex]x_{1} = V_{0}cos \vartheta t[/tex]
[tex]y_{1} = V_{0}sin \vartheta t - \frac{1}{2}gt^2[/tex]
Rock 2:
[tex]x_{2} = d - V_{0}cos \phi t[/tex]
[tex]y_{2} = V_{0}cos \phi t - \frac{1}{2}gt^2[/tex]
[tex] cos \phi = cos(90 - \vartheta) = sin \vartheta[/tex]
[tex] sin \phi = sin (90 - \vartheta) = cos \vartheta[/tex]
[tex]x_{2} - x_{1} = d - V_{0}(cos \vartheta + sin \vartheta)t[/tex]
[tex]y_{2} - y_{1} = V_{0}(cos \vartheta - sin \vartheta)t[/tex]
[tex]Let A = d, B = -V_{0}(cos \vartheta + sin \vartheta) and c = V_{0}(cos \vartheta + sin \vartheta)[/tex]
[tex]D^2 = (A + Bt)^2 + (Ct)^2[/tex]
[tex]= A^2 + 2ABt + B^2t^2 + C^2t^2[/tex]
[tex]= A^2 + 2ABt + (B^2 + C^2)t[/tex]
Recall quadratic equation = ax^2 + bx + c
[tex]Let a = B^2 + C^2, b = 2AB and c = A^2[/tex]
Minimum = vertex
[tex]x = \frac{-b}{2a}[/tex]
[tex]... y_{min} = a(\frac{-b}{2a})^2 + b(\frac{-b}{2a}) + c[/tex]
[tex] = \frac{ab^2}{4a^2} - \frac{b^2}{2a} + c[/tex]
[tex]... d_{min} = \sqrt\frac{4A^2B^2 - 2(4A^2B^2) + A^2}{4(B+C)}[/tex]
[tex] = \sqrt\frac{-A^2B^2 + A^2(B^2 + C^2)}{B^2 + C^2}[/tex]
[tex] = \sqrt\frac{A^2C^2}{B^2 + C^2}[/tex]