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Sagot :
To solve this problem, we need to analyze the motion of the ball under the influence of gravity. Let's go through the steps in detail:
1. Initial Speed: The ball is thrown upward with an initial speed of [tex]\( 5 \, \text{m/s} \)[/tex].
2. Motion Under Gravity: When the ball is thrown upward, gravity acts in the downward direction, decelerating the ball until it comes to a momentary stop at its highest point.
3. Return Journey: After reaching the highest point, the ball will start to fall back towards its starting point. As it falls, it will accelerate due to gravity.
4. Speed at Starting Point: By the time the ball returns to its starting point, it will have gained the same amount of speed it lost while ascending, due to the constant acceleration of gravity. However, the direction of the velocity will be opposite to the direction it was initially thrown.
5. Direction of Velocity: When the ball reaches its starting point, it will be moving downward. Therefore, the speed it has when it returns to the starting point will be the same as the initial speed but in the opposite direction.
Given that the initial speed was [tex]\( 5 \, \text{m/s} \)[/tex] upwards, the speed when it returns to the starting point will be [tex]\( 5 \, \text{m/s} \)[/tex] downwards.
6. Numerical Value: Since we are only interested in the speed, which is always a positive value, the answer is technically [tex]\( 5 \, \text{m/s} \)[/tex]. But if we are considering velocity (which includes direction), we need to acknowledge the change in direction.
Thus, when the ball returns to its starting point, its velocity will be [tex]\( -5 \, \text{m/s} \)[/tex], indicating a downward motion.
In conclusion, the speed of the ball when it returns to its starting point is [tex]\( 5 \, \text{m/s} \)[/tex], and its velocity is [tex]\( -5 \, \text{m/s} \)[/tex].
1. Initial Speed: The ball is thrown upward with an initial speed of [tex]\( 5 \, \text{m/s} \)[/tex].
2. Motion Under Gravity: When the ball is thrown upward, gravity acts in the downward direction, decelerating the ball until it comes to a momentary stop at its highest point.
3. Return Journey: After reaching the highest point, the ball will start to fall back towards its starting point. As it falls, it will accelerate due to gravity.
4. Speed at Starting Point: By the time the ball returns to its starting point, it will have gained the same amount of speed it lost while ascending, due to the constant acceleration of gravity. However, the direction of the velocity will be opposite to the direction it was initially thrown.
5. Direction of Velocity: When the ball reaches its starting point, it will be moving downward. Therefore, the speed it has when it returns to the starting point will be the same as the initial speed but in the opposite direction.
Given that the initial speed was [tex]\( 5 \, \text{m/s} \)[/tex] upwards, the speed when it returns to the starting point will be [tex]\( 5 \, \text{m/s} \)[/tex] downwards.
6. Numerical Value: Since we are only interested in the speed, which is always a positive value, the answer is technically [tex]\( 5 \, \text{m/s} \)[/tex]. But if we are considering velocity (which includes direction), we need to acknowledge the change in direction.
Thus, when the ball returns to its starting point, its velocity will be [tex]\( -5 \, \text{m/s} \)[/tex], indicating a downward motion.
In conclusion, the speed of the ball when it returns to its starting point is [tex]\( 5 \, \text{m/s} \)[/tex], and its velocity is [tex]\( -5 \, \text{m/s} \)[/tex].
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