Answer :
To determine the horizontal displacement ([tex]\(\Delta x\)[/tex]) of a projectile that is launched horizontally, we need to consider the initial conditions and the nature of the motion.
1. Horizontal launch: When a projectile is launched horizontally, it means that its initial velocity has no vertical component. The initial velocity ([tex]\(v_i\)[/tex]) is entirely in the horizontal direction ([tex]\(v_x\)[/tex]).
2. Uniform horizontal velocity: In the absence of horizontal forces, the horizontal component of the velocity remains constant. Therefore, the horizontal displacement depends only on the horizontal component of the initial velocity and the time the projectile is in motion.
3. Time of flight: The time during which the projectile is in the air ([tex]\(\Delta t\)[/tex]) depends on the vertical motion, which is influenced solely by gravity, but this does not directly affect the horizontal displacement formula since the horizontal motion is uniform.
Given this understanding, we look at the formula options provided:
- [tex]\(\Delta x = v_i (\cos \theta)\)[/tex]: This formula accounts for situations when the initial velocity has a component in the direction of [tex]\(\theta\)[/tex]. Since the projectile is launched horizontally ([tex]\(\theta = 0\)[/tex]), the cosine factor simplifies the equation incorrectly for a horizontal launch.
- [tex]\(\Delta x = v_i (\sin \theta) \Delta t\)[/tex]: This formula implies a scenario where the vertical (initially) influences the horizontal, which is not applicable here.
- [tex]\(\Delta x = a_y \Delta t\)[/tex]: This formula incorrectly mixes vertical acceleration ([tex]\(a_y\)[/tex]) and horizontal displacement, which do not correlate directly for horizontal displacement.
- [tex]\(\Delta x = v_x \Delta t\)[/tex]: This is the correct formula. Here, [tex]\(v_x\)[/tex] represents the initial (and constant) horizontal velocity, and [tex]\(\Delta t\)[/tex] is the time the projectile is in the air.
Thus, the proper formula to calculate the horizontal displacement of a horizontally launched projectile is:
[tex]\[ \Delta x = v_x \Delta t \][/tex]
This matches the correct approach for calculating the horizontal displacement given the nature of the motion.
1. Horizontal launch: When a projectile is launched horizontally, it means that its initial velocity has no vertical component. The initial velocity ([tex]\(v_i\)[/tex]) is entirely in the horizontal direction ([tex]\(v_x\)[/tex]).
2. Uniform horizontal velocity: In the absence of horizontal forces, the horizontal component of the velocity remains constant. Therefore, the horizontal displacement depends only on the horizontal component of the initial velocity and the time the projectile is in motion.
3. Time of flight: The time during which the projectile is in the air ([tex]\(\Delta t\)[/tex]) depends on the vertical motion, which is influenced solely by gravity, but this does not directly affect the horizontal displacement formula since the horizontal motion is uniform.
Given this understanding, we look at the formula options provided:
- [tex]\(\Delta x = v_i (\cos \theta)\)[/tex]: This formula accounts for situations when the initial velocity has a component in the direction of [tex]\(\theta\)[/tex]. Since the projectile is launched horizontally ([tex]\(\theta = 0\)[/tex]), the cosine factor simplifies the equation incorrectly for a horizontal launch.
- [tex]\(\Delta x = v_i (\sin \theta) \Delta t\)[/tex]: This formula implies a scenario where the vertical (initially) influences the horizontal, which is not applicable here.
- [tex]\(\Delta x = a_y \Delta t\)[/tex]: This formula incorrectly mixes vertical acceleration ([tex]\(a_y\)[/tex]) and horizontal displacement, which do not correlate directly for horizontal displacement.
- [tex]\(\Delta x = v_x \Delta t\)[/tex]: This is the correct formula. Here, [tex]\(v_x\)[/tex] represents the initial (and constant) horizontal velocity, and [tex]\(\Delta t\)[/tex] is the time the projectile is in the air.
Thus, the proper formula to calculate the horizontal displacement of a horizontally launched projectile is:
[tex]\[ \Delta x = v_x \Delta t \][/tex]
This matches the correct approach for calculating the horizontal displacement given the nature of the motion.
