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Sagot :
To address the question regarding the time it takes for a projectile to go up to its maximum height and the time it takes to return from its maximum height back to its initial height, we need to examine the principles of projectile motion.
Here are the key steps to understand why this statement is true:
1. Symmetry of Projectile Motion:
- When a projectile is launched, it follows a parabolic trajectory under the influence of gravity (assuming no air resistance).
- The motion of the projectile can be divided into two symmetric parts around the peak of its trajectory: the upward motion to reach the maximum height and the downward motion to return to the initial height.
2. Time to Maximum Height:
- When the projectile is launched, it starts with an initial velocity [tex]\( v_0 \)[/tex] at an angle [tex]\( \theta \)[/tex] to the horizontal.
- The vertical component of the initial velocity is given by [tex]\( v_{0y} = v_0 \sin(\theta) \)[/tex].
- Using the kinematic equations, the time [tex]\( t_{\text{up}} \)[/tex] taken to reach maximum height can be found using:
[tex]\[ v_{0y} = g \cdot t_{\text{up}} \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity.
- Solving for [tex]\( t_{\text{up}} \)[/tex], we get:
[tex]\[ t_{\text{up}} = \frac{v_{0y}}{g} \][/tex]
3. Time from Maximum Height to Initial Height:
- At the maximum height, the projectile temporarily has zero vertical velocity.
- Gravity acts on the projectile, causing it to accelerate downwards at [tex]\( g \)[/tex].
- The time taken [tex]\( t_{\text{down}} \)[/tex] for the projectile to descend from the maximum height back to the initial height is the same as the time taken to ascend because the motion is symmetric.
- Thus:
[tex]\[ t_{\text{down}} = \frac{v_{0y}}{g} \][/tex]
4. Total Time of Flight:
- The total time of flight is the sum of [tex]\( t_{\text{up}} \)[/tex] and [tex]\( t_{\text{down}} \)[/tex]:
[tex]\[ t_{\text{total}} = t_{\text{up}} + t_{\text{down}} = \frac{v_{0y}}{g} + \frac{v_{0y}}{g} = 2 \cdot \frac{v_{0y}}{g} \][/tex]
Hence, the time it takes for a projectile to reach its maximum height is exactly equal to the time it takes to fall back down to its initial height because they are part of the same symmetric motion under gravity.
Therefore, the statement "The time it takes for a projectile to go up to its maximum height is the same amount of time that the projectile takes to fall from its maximum height back down to its initial height." is True.
Here are the key steps to understand why this statement is true:
1. Symmetry of Projectile Motion:
- When a projectile is launched, it follows a parabolic trajectory under the influence of gravity (assuming no air resistance).
- The motion of the projectile can be divided into two symmetric parts around the peak of its trajectory: the upward motion to reach the maximum height and the downward motion to return to the initial height.
2. Time to Maximum Height:
- When the projectile is launched, it starts with an initial velocity [tex]\( v_0 \)[/tex] at an angle [tex]\( \theta \)[/tex] to the horizontal.
- The vertical component of the initial velocity is given by [tex]\( v_{0y} = v_0 \sin(\theta) \)[/tex].
- Using the kinematic equations, the time [tex]\( t_{\text{up}} \)[/tex] taken to reach maximum height can be found using:
[tex]\[ v_{0y} = g \cdot t_{\text{up}} \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity.
- Solving for [tex]\( t_{\text{up}} \)[/tex], we get:
[tex]\[ t_{\text{up}} = \frac{v_{0y}}{g} \][/tex]
3. Time from Maximum Height to Initial Height:
- At the maximum height, the projectile temporarily has zero vertical velocity.
- Gravity acts on the projectile, causing it to accelerate downwards at [tex]\( g \)[/tex].
- The time taken [tex]\( t_{\text{down}} \)[/tex] for the projectile to descend from the maximum height back to the initial height is the same as the time taken to ascend because the motion is symmetric.
- Thus:
[tex]\[ t_{\text{down}} = \frac{v_{0y}}{g} \][/tex]
4. Total Time of Flight:
- The total time of flight is the sum of [tex]\( t_{\text{up}} \)[/tex] and [tex]\( t_{\text{down}} \)[/tex]:
[tex]\[ t_{\text{total}} = t_{\text{up}} + t_{\text{down}} = \frac{v_{0y}}{g} + \frac{v_{0y}}{g} = 2 \cdot \frac{v_{0y}}{g} \][/tex]
Hence, the time it takes for a projectile to reach its maximum height is exactly equal to the time it takes to fall back down to its initial height because they are part of the same symmetric motion under gravity.
Therefore, the statement "The time it takes for a projectile to go up to its maximum height is the same amount of time that the projectile takes to fall from its maximum height back down to its initial height." is True.
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