To solve this problem, we need to break it down into two parts: the vertical motion and the horizontal motion.

1. Vertical Motion:
First, we need to find the time it takes for the rock to reach the window. We know the initial vertical velocity (Viy), the final height (y), and the initial height (yi), and we know that the acceleration due to gravity (a) is -9.8 m/s² (negative because it's downward). We can use the following kinematic equation to solve for time (t):

y = yi + Viy*t + 0.5*a*t²

Rearranging for t gives us:

t = [y - yi - Viy*t] / [0.5*a]

We know that the initial vertical velocity Viy can be found using the equation Viy = Vi*sin(θ), where Vi is the initial speed of the rock and θ is the angle at which it was thrown. So, Viy = 30.0 m/s * sin(40.0°) = 19.28 m/s.

Substituting these values into the equation gives us:

t = [18.0 m - 2.0 m - 19.28 m/s*t] / [0.5*-9.8 m/s²]
Solving this equation gives us t = 1.03 s.

2. Horizontal Motion:
Now that we have the time, we can find the horizontal distance (x) the rock traveled. We know the initial horizontal velocity (Vix) and that the horizontal acceleration (ax) is 0 (because we're ignoring air resistance). We can use the following kinematic equation:

x = xi + Vix*t + 0.5*ax*t²

Since the rock was released from rest, xi = 0, and since there's no horizontal acceleration, ax = 0. So the equation simplifies to:

x = Vix*t

We know that the initial horizontal velocity Vix can be found using the equation Vix = Vi*cos(θ). So, Vix = 30.0 m/s * cos(40.0°) = 22.94 m/s.

Substituting these values into the equation gives us:

x = 22.94 m/s * 1.03 s = 23.63 m

So, the rock was released from a horizontal distance of approximately 23.63 meters from the window.

Question

To solve this problem, we need to break it down into two parts: the vertical motion and the horizontal motion.

1. Vertical Motion: 
First, we need to find the time it takes for the rock to reach the window. We know the initial vertical velocity (Viy), the final height (y), and the initial height (yi), and we know that the acceleration due to gravity (a) is -9.8 m/s² (negative because it's downward). We can use the following kinematic equation to solve for time (t):

y = yi + Viy*t + 0.5*a*t²

Rearranging for t gives us:

t = [y - yi - Viy*t] / [0.5*a]

We know that the initial vertical velocity Viy can be found using the equation Viy = Vi*sin(θ), where Vi is the initial speed of the rock and θ is the angle at which it was thrown. So, Viy = 30.0 m/s * sin(40.0°) = 19.28 m/s.

Substituting these values into the equation gives us:

t = [18.0 m - 2.0 m - 19.28 m/s*t] / [0.5*-9.8 m/s²]
Solving this equation gives us t = 1.03 s.

2. Horizontal Motion: 
Now that we have the time, we can find the horizontal distance (x) the rock traveled. We know the initial horizontal velocity (Vix) and that the horizontal acceleration (ax) is 0 (because we're ignoring air resistance). We can use the following kinematic equation:

x = xi + Vix*t + 0.5*ax*t²

Since the rock was released from rest, xi = 0, and since there's no horizontal acceleration, ax = 0. So the equation simplifies to:

x = Vix*t

We know that the initial horizontal velocity Vix can be found using the equation Vix = Vi*cos(θ). So, Vix = 30.0 m/s * cos(40.0°) = 22.94 m/s.

Substituting these values into the equation gives us:

x = 22.94 m/s * 1.03 s = 23.63 m

So, the rock was released from a horizontal distance of approximately 23.63 meters from the window.

Knowee AI · Accepted Answer

To solve this problem, we need to break it down into two parts: the vertical motion and the horizontal motion.

1. Vertical Motion: 
First, we need to find the time it takes for the rock to reach the window. We know the initial vertical velocity (Viy), the final height (y), and the initial height (yi), and we know that the acceleration due to gravity (a) is -9.8 m/s² (negative because it's downward). We can use the following kinematic equation to solve for time (t):

y = yi + Viy*t + 0.5*a*t²

Rearranging for t gives us:

t = [y - yi - Viy*t] / [0.5*a]

We know that the initial vertical velocity Viy can be found using the equation Viy = Vi*sin(θ), where Vi is the initial speed of the rock and θ is the angle at which it was thrown. So, Viy = 30.0 m/s * sin(40.0°) = 19.28 m/s.

Substituting these values into the equation gives us:

t = [18.0 m - 2.0 m - 19.28 m/s*t] / [0.5*-9.8 m/s²]
Solving this equation gives us t = 1.03 s.

2. Horizontal Motion: 
Now that we have the time, we can find the horizontal distance (x) the rock traveled. We know the initial horizontal velocity (Vix) and that the horizontal acceleration (ax) is 0 (because we're ignoring air resistance). We can use the following kinematic equation:

x = xi + Vix*t + 0.5*ax*t²

Since the rock was released from rest, xi = 0, and since there's no horizontal acceleration, ax = 0. So the equation simplifies to:

x = Vix*t

We know that the initial horizontal velocity Vix can be found using the equation Vix = Vi*cos(θ). So, Vix = 30.0 m/s * cos(40.0°) = 22.94 m/s.

Substituting these values into the equation gives us:

x = 22.94 m/s * 1.03 s = 23.63 m

So, the rock was released from a horizontal distance of approximately 23.63 meters from the window.

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