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Sagot :
Explications:
To solve this problem, we can use the equations of motion to find the time it takes for the balls to collide. Then, we can use the time to determine the velocities of projection for both balls.
Let's denote the initial velocity of projection of ball A as \( v_{A} \) and that of ball B as \( v_{B} \).
The horizontal and vertical components of the initial velocity of ball A are:
- \( v_{A_x} = v_{A} \cos(30^\circ) \)
- \( v_{A_y} = v_{A} \sin(30^\circ) \)
Similarly, for ball B:
- \( v_{B_x} = v_{B} \cos(120^\circ) \)
- \( v_{B_y} = v_{B} \sin(120^\circ) \)
The horizontal distance traveled by both balls will be the same when they collide, so we can set up the equation:
\[ v_{A_x} \cdot t = v_{B_x} \cdot t \]
\[ v_{A} \cdot \cos(30^\circ) \cdot t = v_{B} \cdot \cos(120^\circ) \cdot t \]
Canceling out the \( t \) terms, we get:
\[ v_{A} \cdot \cos(30^\circ) = v_{B} \cdot \cos(120^\circ) \]
Similarly, we can set up an equation using the vertical components:
\[ v_{A_y} \cdot t + \frac{1}{2} \cdot (-9.8) \cdot t^2 = v_{B_y} \cdot t + \frac{1}{2} \cdot (-9.8) \cdot t^2 \]
\[ v_{A} \cdot \sin(30^\circ) \cdot t = v_{B} \cdot \sin(120^\circ) \cdot t \]
Canceling out the \( t \) terms, we get:
\[ v_{A} \cdot \sin(30^\circ) = v_{B} \cdot \sin(120^\circ) \]
Now, we can divide the first equation by the second equation to eliminate \( t \) and solve for the ratio of velocities:
\[ \frac{v_{A} \cdot \cos(30^\circ)}{v_{A} \cdot \sin(30^\circ)} = \frac{v_{B} \cdot \cos(120^\circ)}{v_{B} \cdot \sin(120^\circ)} \]
\[ \frac{\cos(30^\circ)}{\sin(30^\circ)} = \frac{\cos(120^\circ)}{\sin(120^\circ)} \]
\[ \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \]
\[ \sqrt{3} = -\frac{1}{\sqrt{3}} \]
\[ \frac{v_{A}}{v_{B}} = \frac{-1}{\sqrt{3}} \]
However, velocity cannot be negative, so the ratio of the velocity of projection of A to that of B is \( \frac{1}{\sqrt{3}} \).
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