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Sagot :
To calculate the horizontal displacement of a horizontally launched projectile, we need to consider the horizontal component of the projectile's motion. For a horizontally launched projectile, the initial velocity is entirely in the horizontal direction, and there is no initial vertical component.
Given the choices:
1. [tex]\(\Delta x = v_i (\cos \theta)\)[/tex]
2. [tex]\(\Delta x = v_i (\sin \theta) \Delta t\)[/tex]
3. [tex]\(\Delta x = a_y \Delta t\)[/tex]
4. [tex]\(\Delta x = v_x \Delta t\)[/tex]
Let's evaluate them:
1. [tex]\(\Delta x = v_i (\cos \theta)\)[/tex]: This expression involves [tex]\(\cos \theta\)[/tex], which is used when an initial velocity has both horizontal and vertical components (i.e., when the launch angle [tex]\(\theta\)[/tex] is not zero). Since the projectile is launched horizontally, [tex]\(\theta = 0\)[/tex], so [tex]\(\cos \theta = 1\)[/tex], making [tex]\(\Delta x = v_i\)[/tex]. However, this does not involve time and is not a complete representation in the context of displacement over time.
2. [tex]\(\Delta x = v_i (\sin \theta) \Delta t\)[/tex]: This expression involves [tex]\(\sin \theta\)[/tex], which is used for the vertical component of the initial velocity when the projectile is launched at an angle. Since the projectile is launched horizontally, [tex]\(\sin \theta = 0\)[/tex], making the displacement zero. This does not correctly represent the horizontal displacement.
3. [tex]\(\Delta x = a_y \Delta t\)[/tex]: This expression suggests that horizontal displacement is a product of vertical acceleration ([tex]\(a_y\)[/tex]) and time. Vertical acceleration due to gravity does not affect horizontal displacement directly, so this formula is incorrect for horizontal displacement.
4. [tex]\(\Delta x = v_x \Delta t\)[/tex]: This expression correctly represents horizontal displacement. [tex]\(v_x\)[/tex] is the horizontal component of the initial velocity, and [tex]\(\Delta t\)[/tex] is the time for which the projectile is in motion. The horizontal displacement ([tex]\(\Delta x\)[/tex]) is directly proportional to the horizontal velocity ([tex]\(v_x\)[/tex]) and the time ([tex]\(\Delta t\)[/tex]).
Hence, the correct formula to calculate the horizontal displacement of a horizontally launched projectile is:
[tex]\[ \Delta x = v_x \Delta t \][/tex]
The correct choice from the given options is:
[tex]\[ \Delta x = v_x \Delta t \][/tex]
Given the choices:
1. [tex]\(\Delta x = v_i (\cos \theta)\)[/tex]
2. [tex]\(\Delta x = v_i (\sin \theta) \Delta t\)[/tex]
3. [tex]\(\Delta x = a_y \Delta t\)[/tex]
4. [tex]\(\Delta x = v_x \Delta t\)[/tex]
Let's evaluate them:
1. [tex]\(\Delta x = v_i (\cos \theta)\)[/tex]: This expression involves [tex]\(\cos \theta\)[/tex], which is used when an initial velocity has both horizontal and vertical components (i.e., when the launch angle [tex]\(\theta\)[/tex] is not zero). Since the projectile is launched horizontally, [tex]\(\theta = 0\)[/tex], so [tex]\(\cos \theta = 1\)[/tex], making [tex]\(\Delta x = v_i\)[/tex]. However, this does not involve time and is not a complete representation in the context of displacement over time.
2. [tex]\(\Delta x = v_i (\sin \theta) \Delta t\)[/tex]: This expression involves [tex]\(\sin \theta\)[/tex], which is used for the vertical component of the initial velocity when the projectile is launched at an angle. Since the projectile is launched horizontally, [tex]\(\sin \theta = 0\)[/tex], making the displacement zero. This does not correctly represent the horizontal displacement.
3. [tex]\(\Delta x = a_y \Delta t\)[/tex]: This expression suggests that horizontal displacement is a product of vertical acceleration ([tex]\(a_y\)[/tex]) and time. Vertical acceleration due to gravity does not affect horizontal displacement directly, so this formula is incorrect for horizontal displacement.
4. [tex]\(\Delta x = v_x \Delta t\)[/tex]: This expression correctly represents horizontal displacement. [tex]\(v_x\)[/tex] is the horizontal component of the initial velocity, and [tex]\(\Delta t\)[/tex] is the time for which the projectile is in motion. The horizontal displacement ([tex]\(\Delta x\)[/tex]) is directly proportional to the horizontal velocity ([tex]\(v_x\)[/tex]) and the time ([tex]\(\Delta t\)[/tex]).
Hence, the correct formula to calculate the horizontal displacement of a horizontally launched projectile is:
[tex]\[ \Delta x = v_x \Delta t \][/tex]
The correct choice from the given options is:
[tex]\[ \Delta x = v_x \Delta t \][/tex]
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