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Horizontal displacement in projectile motion is the distance traveled in the horizontal direction by an object. It is defined by the formula: \[ x = v_{0x} t \] where \( v_{0x} \) is the horizontal component of the initial velocity and \( t \) is the time elapsed. Since there is no horizontal acceleration in projectile motion, the horizontal velocity remains constant throughout the motion. This means that you simply multiply the horizontal component of the initial velocity by the time to find the horizontal distance traveled.

• For time \( t=1.10 \) seconds, the horizontal component is calculated as \( 13.79 \text{ m/s} \, \times\, 1.10 \text{ s} \approx 15.17 \text{ m} \).
• For time \( t=1.80 \) seconds, it becomes \( 13.79 \text{ m/s} \, \times\, 1.80 \text{ s} \approx 24.82 \text{ m} \).
• At \( t=5.00 \) seconds, the displacement is \( 13.79 \text{ m/s} \, \times\, 5.00 \text{ s} \approx 68.95 \text{ m} \).

As you can see, the horizontal displacement is solely dependent on time and the initial horizontal velocity component. It's important to note that without any additional forces acting in the horizontal plane, the velocity remains constant.

Vertical displacement involves the upward or downward distance traveled by an object, under the influence of gravity, in projectile motion. Vertical motion is influenced by gravitational acceleration denoted by \( g \), which is \( 9.8 \text{ m/s}^2 \) downward.For vertical displacement, the formula to use is:\[ y = v_{0y} t - \frac{1}{2} g t^2 \] where \( v_{0y} \) is the vertical component of the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time elapsed. Gravity affects the motion, causing the object to decelerate upwards and accelerate downwards.

• At \( t=1.10 \) seconds, the calculation is \( 11.57 \times 1.10 - \frac{1}{2} \times 9.8 \times (1.10)^2 \approx 4.48 \text{ m} \).
• For \( t=1.80 \) seconds, it's \( 11.57 \times 1.80 - \frac{1}{2} \times 9.8 \times (1.80)^2 \approx 5.98 \text{ m} \).
• At \( t=5.00 \) seconds, the vertical displacement changes significantly due to the longer time period and gravity's effect: \( 11.57 \times 5.00 - \frac{1}{2} \times 9.8 \times (5.00)^2 \approx -56.05 \text{ m} \). This negative value indicates the object has fallen below the original launch height.

Initial velocity is crucial in defining the motion of a projectile. It consists of two parts: the horizontal and vertical components, expressed in terms of magnitude and angle. When an object is propelled, its initial velocity is resolved into these two perpendicular components using trigonometric functions based on its launch angle.For a projectile launched with an initial velocity \( v_0 \) at an angle \( \theta \) from the horizontal, the components are:\[ v_{0x} = v_0 \cos(\theta) \] \[ v_{0y} = v_0 \sin(\theta) \] These equations are derived from basic trigonometry:

• The horizontal component, \( v_{0x} \), reflects the speed at which the projectile travels horizontally.
• The vertical component, \( v_{0y} \), shows how fast it moves upwards against gravity initially.

In the example of our catapulted stone:

• The initial velocity \( v_0 = 18.0 \text{ m/s} \) and the launch angle \( \theta = 40.0^{\circ} \).
• Using the equations above, we find \( v_{0x} = 18.0 \cos(40.0^{\circ}) \approx 13.79 \text{ m/s} \).
• Similarly, \( v_{0y} = 18.0 \sin(40.0^{\circ}) \approx 11.57 \text{ m/s} \).

Understanding the breakdown of initial velocity into these components helps in predicting the projectile's trajectory and determining both horizontal and vertical displacements at any given time.