Answer :
To determine the correct formula for calculating the horizontal displacement of a horizontally launched projectile, we'll explore each option provided:
1. [tex]\(\Delta x = v (\cos \theta)\)[/tex]:
- This formula considers the horizontal component of velocity for an angle [tex]\(\theta\)[/tex]. However, in a horizontally launched projectile, the launch angle [tex]\(\theta\)[/tex] is zero, making [tex]\(\cos \theta = 1\)[/tex]. Therefore, it simplifies to [tex]\(\Delta x = v\)[/tex], which isn't enough since time is not considered directly.
2. [tex]\(\Delta x = v (\sin \theta) \Delta t\)[/tex]:
- This formula relates to the vertical component of velocity since sine of the angle is used. But, horizontal displacement depends only on the horizontal velocity component, and this isn't relevant for horizontal displacement of a projectile.
3. [tex]\(\Delta x = a \Delta t\)[/tex]:
- This is not accurate, as it suggests displacement results from acceleration directly affecting time without proper integration over time leading to squared time terms and missing initial velocity.
4. [tex]\(\Delta x = v_x \Delta t\)[/tex]:
- This formula correctly represents the horizontal displacement of a horizontally launched projectile. Here, [tex]\(v_x\)[/tex] is the horizontal component of the velocity (which is the initial velocity in horizontal launch as there is no vertical initial component), and [tex]\(\Delta t\)[/tex] is the time of travel. This matches our need as horizontal displacement is simply the constant horizontal velocity multiplied by the time of travel.
Given these points, the correct answer is:
[tex]\(\Delta x = v_x \Delta t\)[/tex].
Thus, the choice is [tex]\( \boxed{4} \)[/tex].
1. [tex]\(\Delta x = v (\cos \theta)\)[/tex]:
- This formula considers the horizontal component of velocity for an angle [tex]\(\theta\)[/tex]. However, in a horizontally launched projectile, the launch angle [tex]\(\theta\)[/tex] is zero, making [tex]\(\cos \theta = 1\)[/tex]. Therefore, it simplifies to [tex]\(\Delta x = v\)[/tex], which isn't enough since time is not considered directly.
2. [tex]\(\Delta x = v (\sin \theta) \Delta t\)[/tex]:
- This formula relates to the vertical component of velocity since sine of the angle is used. But, horizontal displacement depends only on the horizontal velocity component, and this isn't relevant for horizontal displacement of a projectile.
3. [tex]\(\Delta x = a \Delta t\)[/tex]:
- This is not accurate, as it suggests displacement results from acceleration directly affecting time without proper integration over time leading to squared time terms and missing initial velocity.
4. [tex]\(\Delta x = v_x \Delta t\)[/tex]:
- This formula correctly represents the horizontal displacement of a horizontally launched projectile. Here, [tex]\(v_x\)[/tex] is the horizontal component of the velocity (which is the initial velocity in horizontal launch as there is no vertical initial component), and [tex]\(\Delta t\)[/tex] is the time of travel. This matches our need as horizontal displacement is simply the constant horizontal velocity multiplied by the time of travel.
Given these points, the correct answer is:
[tex]\(\Delta x = v_x \Delta t\)[/tex].
Thus, the choice is [tex]\( \boxed{4} \)[/tex].
