he exit velocity of water from a commercially available water cannon that can be mounted on a truck is about 100.7 m/s* If we ignore air resistance, how far away would the water protectors at Standing Rock have to stand in order not to get sprayed with this water cannon?***The two pictures are explanations and answers, and I hope you solve the problem like this way. Please provide a detailed explanation and solution. 1. **Projectile Motion and Range Calculation for a Water Cannon** The exit velocity of water from a commercially available water cannon, which can be mounted on a truck, is approximately 100.7 m/s. Ignoring air resistance, we aim to determine how far a person would need to stand to avoid getting sprayed by this water cannon at Standing Rock. During our projectile motion lab, we determined that a launch angle of 45° maximizes the range of a projectile when launched from ground level over flat terrain. Below is the step-by-step procedure to calculate the range: **Coordinate System for Projectile Motion:** - We draw a diagram and label a coordinate system with `x = 0` at the launch point and `y = 0` on the ground. **Key Equations and Concepts:** - The horizontal and vertical motions are treated separately: - Horizontal motion is a constant velocity problem. - Vertical motion is a constant acceleration problem. - Kinematic equations used are: - \( x_f = v_{i,x} t_f \) - \( y_f = 0 = v_{i,y} t_f - \frac{1}{2} g t_f^2 \) **Breaking Initial Velocity into Components:** - The launch velocity (\( v_i \)) is split into x- and y-components using trigonometry: - \( v_{i,x} = v_i \cos 45^\circ \) - \( v_{i,y} = v_i \sin 45^\circ \) **Calculations Involving Trigonometry:** - Using the angle of 45°, the trigonometric identities are: - \( \sin 45^\circ = \frac{v_{i,y}}{v_i} \rightarrow v_{i,y} = v_i \sin 45^\circ \) - \( \cos 45^\circ = \frac{v_{i,x}}{v_i} \rightarrow v_{i,x} = v_i \cos 45^\circ \) **Solving the Equations:** - Plug the velocity components into the kinematic equations: - For horizontal: \( x_f = (v_i \cos 45^\circ) t_f \) - For vertical (simplified for \( t_f \)): \( 0 = (v_i \sin 45 The image shows a step-by-step solution for calculating the horizontal distance (\(x_f\)) using projectile motion equations.1. **Equation Derivation**: - The time of flight (\(t_f\)) is derived from the equation: \[ v_i \sin 45^\circ - \frac{1}{2}gt_f = 0 \] Simplifying gives: \[ t_f = \frac{2v_i \sin 45^\circ}{g} \]2. **First Equation**: - The horizontal distance \(x_f\) is given by: \[ x_f = (v_i \cos 45^\circ) t_f = \frac{2}{g} v_i^2 \cos 45^\circ \sin 45^\circ \] - Further simplifies to: \[ = \frac{2}{g} v_i^2 \left(\frac{1}{2}\right) \]3. **Calculations**: - Substituting \(v_i = 100.7 \, \text{m/s}\) and \(g = 9.81 \, \text{m/s}^2\), the calculation becomes: \[ x_f = \frac{2}{9.81} (100.7 \, \text{m/s})^2 \left(\frac{1}{2}\right) = 1,033.689 \, \text{m} \]4. **Rounding**: - Rounded to **three significant figures**: \(x_f \approx 1,030 \, \text{m}\). - Calculation repeated with \(g = 9.80 \, \text{m/s}^2\) yielding the same rounded value. - For \(g = 9.8 \, \text{m/s}^2\), rounding to **two significant figures** gives \(x_f \approx 1.0 \times 10^3 \, \text{m}\).This solution methodically uses trigonometric components and includes detailed rounding based on significant figures.

1. The exit velocity of water from a commercially available water cannon that can be mounted on a truck is about 100.7 m/s* If we ignore air resistance, how far away would the water protectors at Standing Rock have to stand in order not to get sprayed with this water cannon? We found in the projectile motion lab that a launch angle of 45° maximizes the range of a projectile launched from ground level over flat ground. We draw a diagram then choose and label a coordinate system: Remember that we can treat the vertical motion and the horizontal motion separately–determining the vertical motion is a constant acceleration problem and determining the horizontal motion is a constant velocity problem. For each direction we'll use the kinematic equation for distance: Xf = *; + Vixtg → X; = Vixtj 3y = 9, + Vayty – 5gt;² → 0 = v.ytg –59t; y=O X= 0

1. The exit velocity of water from a commercially available water cannon that can be mounted on a truck is about 100.7 m/s* If we ignore air resistance, how far away would the water protectors at Standing Rock have to stand in order not to get sprayed with this water cannon? We found in the projectile motion lab that a launch angle of 45° maximizes the range of a projectile launched from ground level over flat ground. We draw a diagram then choose and label a coordinate system: Remember that we can treat the vertical motion and the horizontal motion separately–determining the vertical motion is a constant acceleration problem and determining the horizontal motion is a constant velocity problem. For each direction we'll use the kinematic equation for distance: Xf = *; + Vixtg → X; = Vixtj 3y = 9, + Vayty – 5gt;² → 0 = v.ytg –59t; y=O X= 0