| 1 $\theta$ could be arbitrary | 0.40 |
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2
The formula for the range \[R = \frac{v_0^2 \sin 2 \alpha}{g}\] of a projectile launched with initial speed $v_0$ at angle $\alpha$ above the horizontal is used |
0.50 |
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| 3 Obtained $\alpha = \pi/2 - \theta$ | 0.20 |
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4
Obtained \[v_0 = \sqrt{\frac{Dg}{\sin 2\theta}}\] |
0.40 |
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1
Condition of equal range is written: \[ \sin (\pi - 2 \theta + \phi ) = \sin (\pi - 2 \theta - \phi) \] |
0.80 |
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| 2 Obtained $\theta=\pi/4$ | 0.40 |
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| 3 Specified that $\phi$ can be arbitrary | 0.40 |
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| 4 \[v_0 = \sqrt{\frac{Dg}{\cos \phi}}\] | 0.40 |
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| 1 Distance between impact points is $R \sqrt{2}$ | 0.20 |
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2
Vertical motion is described \[ 2v_0 \sin \left( \frac{\pi}{4} + \frac{\phi}{2} \right) = g t_1 \] |
0.50 |
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3
Horizontal motion is described \[ v_0 \cos \left( \frac{\pi}{4} + \frac{\phi}{2} \right) t_1 = R \sqrt{2} \] |
0.50 |
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4
Trigonometry is used: \[\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\]or \[\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\]or \[ \sin \left( \frac{\pi}{4} + \frac{\phi}{2} \right) = \cos\left( \frac{\pi}{4} - \frac{\phi}{2} \right)\] |
1.00 |
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| 5 \[R = \frac{gt_1 t_2}{2\sqrt{2}}\] | 0.70 |
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| 6 It is written that value is unique | 0.10 |
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1
It's used that \[ \Delta x = \frac{gt_1t_2}{2}\] |
0.30 |
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| 2 \[y'(x) = -2 \cos \frac{2x}{L}\] | 0.50 |
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| 3 In impact points $y' = \pm 1$ | 0.30 |
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| 4 The first case is found: left impact point $2x/L=\pi/3$, right impact point $2x/L=2\pi/3$ | 0.70 |
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5
Answer for the first case: \[L=\frac{3gt_1 t_2}{\pi}\] |
0.50 |
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| 6 The second case is found: left impact point $2x/L=-\pi/3$, right impact point $2x/L=4\pi/3$ | 0.70 |
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7
Answer for the second case: \[L=\frac{3gt_1 t_2}{5 \pi}\] |
0.50 |
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