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Projectile motion refers to the path that an object follows when it is launched into the air with an initial velocity, and it moves under the influence of gravity. In our water gun scenario, the water behaves as a projectile. This means as soon as it leaves the muzzle, it traces a curved path due to gravity pulling it downwards while it moves horizontally.Key features of projectile motion include:

• Two Components of Motion: The motion of the water has two components - horizontal and vertical. The horizontal component is constant because we ignore air resistance, whereas the vertical component is affected by gravity.
• Time of Flight: The time taken for the projectile to reach the ground is determined by the vertical motion. Here, we use the formula: \( \Delta y = v_{i_y} t + \frac{1}{2} g t^2 \), where \( \Delta y \) is the vertical displacement, \( v_{i_y} = 0 \) is the initial vertical velocity, and \( g = 9.81 \, \text{m/s}^2 \) is gravitational acceleration.Rearranging the equation allows us to solve for time \( t \), which in this scenario is approximately 0.39 seconds.
• Horizontal Range: The horizontal distance traveled by the projectile (here, water) can be calculated by multiplying the horizontal velocity by the time of flight.

Gauge pressure is an important concept in fluid dynamics. It refers to the pressure inside a system relative to the atmospheric pressure outside the system. When dealing with the water gun, gauge pressure indicates the additional pressure exerted by the water due to pumping, beyond what is present naturally in the environment.Calculating Gauge Pressure:
For this problem, we employed Bernoulli's principle, which connects the speed of a fluid and its pressure. Bernoulli's equation is:\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]Given that the initial speed of the water in the reservoir \( v_1 = 0 \) and the system is held at a height where the vertical distance \( h_2 = 0.75 \, \text{m} \), we can simplify to:\[ P_{gauge} = \frac{1}{2} (1000 \, \text{kg/m}^3) (v_x^2) - (1000 \, \text{kg/m}^3) (9.81 \, \text{m/s}^2) (0.75 \, \text{m}) \]Here, \( v_x \) is the horizontal velocity computed, and 169100 Pa is calculated as the gauge pressure. This value indicates how much the water has been pressurized by the pumping action beyond atmospheric levels.

Horizontal velocity explores the component of a projectile's velocity that remains constant (in the absence of air resistance) throughout its motion. For our water gun scenario, determining the horizontal velocity is key to solving the problem of where and how far the water will land.Understanding Horizontal Velocity:
In the case of the water gun, the horizontal velocity, \( v_x \), can be calculated once we know the time the projectile is in the air. We use the formula for horizontal motion:\[ v_x = \frac{\text{horizontal distance}}{\text{time}} \]Applying this calculation to our problem:

• We found the time of flight to be 0.39 seconds through vertical motion equations.
• The horizontal distance travelled by the water was measured as 7.3 meters.

Substituting these values gives us:\[ v_x = \frac{7.3 \, \text{m}}{0.39 \, \text{s}} \approx 18.72 \, \text{m/s} \]This velocity indicates how fast the water shoots out from the gun horizontally, a crucial factor in applying Bernoulli's equation to find the gauge pressure in the water gun's reservoir.