Problem 69 - University Physics with Modern Physics

Question

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of your rich uncle. The top of the fence is 5.00 m above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60 m above the ground and at an angle of 56.0\(^\circ\) above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance beyond the fence will the rock land on the ground?

Step-by-Step Solution

Answer

Minimum speed: 15.1 m/s. Horizontal distance beyond the fence: 26.4 m.

Step 1: Analyze Given Data

First, let's review the given data: - Horizontal distance to the fence, \( x_0 = 14.0 \) m.- Height of the fence, \( h_f = 5.0 \) m.- Initial height of the rock, \( h_0 = 1.60 \) m.- Angle of projection, \( \theta = 56.0^\circ \).- Gravitational acceleration, \( g = 9.81 \text{ m/s}^2 \).

Step 2: Establish the Equation for Vertical Motion

To find the minimum speed, we need to determine the time \( t \) it takes for the rock to reach the height of the fence using the vertical motion equation:\[h = h_0 + v_0 \sin(\theta) t - \frac{1}{2}gt^2\]Here, \( h = 5.0 \text{ m} \).

Step 3: Derive the Horizontal Motion Equation

The horizontal motion equation is given by:\[x = x_0 + v_0 \cos(\theta) t\]Here, \( x = 14.0 \text{ m} \), and this solves for the time \( t \) which can be derived from the horizontal distance.

Step 4: Solve the System of Equations

Using Cartesian equations for both motions, set up the system of equations from Steps 2 and 3. The solution for both equations will provide the potential initial velocity \( v_0 \). This yields a quadratic equation in terms of \( v_0 \).

Step 5: Solve for Minimum Initial Speed

Solve the quadratic equation for \( v_0 \) to find the minimum speed required to just clear the fence:\[v_0 = \left(\frac{g x^2}{2 (h - h_0 - x \tan(\theta)) \cos^2(\theta) }\right)^{1/2}\]Calculate with given values to find \( v_0 \).

Step 6: Equation for Horizontal Distance Beyond Fence

Now use the velocity obtained to find the time it takes for the rock to fall to the ground from height 1.60 m after clearing the fence:\[0 = h_f + v_0 \sin(\theta)t' - \frac{1}{2}gt'^2 \]Here \( t' \) is the time after the rock clears the fence.

Step 7: Calculate the Total Horizontal Distance

Use the total time \( t + t' \) to calculate how far the rock travels horizontally beyond the fence:\[x_b = v_0 \cos(\theta)(t + t') - x_0\]Calculate to find \( x_b \).

Key Concepts

Physics Problem

Projectile motion problems in physics often involve predicting the path of an object that is launched with an initial velocity at an angle to the horizontal. In the given exercise, your goal is to determine the minimum initial speed required to throw a rock over a fence and to find out where the rock will land beyond the fence.

• The rock starts at a height of 1.60 m above ground.
• The fence, which needs to be cleared, is 5.00 m high.
• The rock is thrown from a distance of 14.0 m away from the fence.
• The throw angle is 56.0° above the horizontal.

This setup reveals a kinematics problem involving vertical and horizontal motion, where you need to calculate the minimum initial velocity and the resulting trajectory of the rock.

Kinematics

Understanding kinematics is essential for solving projectile motion problems. Kinematics deals with the analysis of motion, excluding the forces that cause it. This domain of physics uses equations to describe the motion of objects using variables such as displacement, velocity, time, and acceleration.In this particular exercise:
1. **Vertical Motion**: The rock's vertical motion is influenced by gravity, reducing its upwards velocity till it reaches the peak, after which it descends under gravity’s influence.
2. **Horizontal Motion**: The horizontal distance covered by the rock is analyzed using a constant horizontal velocity, as horizontal motion under projectile conditions is unaffected by gravity.Key kinematic equations help in determining both vertical and horizontal positions at any given time. For instance, the vertical motion can be characterized by the equation:\[h = h_0 + v_0 \sin(\theta) t - \frac{1}{2}gt^2\]And for horizontal motion, the distance is given by:\[x = x_0 + v_0 \cos(\theta) t\]These equations are interdependent and need to be solved simultaneously in projectile problems to find unknowns like time and initial velocity.

Motion Equations

The solution to projectile motion problems like this one requires setting up and solving motion equations for both vertical and horizontal components. These equations allow you to find necessary parameters, such as initial velocity and range.### Vertical Motion EquationFor vertical motion, the equation accounts for initial height, vertical velocity component, and acceleration due to gravity:\[h = h_0 + v_0 \sin(\theta) t - \frac{1}{2}gt^2\]This helps in determining the time taken to reach a specific vertical height, such as the top of the fence.### Horizontal Motion EquationHorizontal motion is addressed by:\[x = x_0 + v_0 \cos(\theta) t\]This equation gives the horizontal component of velocity, allowing calculation of the time required to reach the fence and beyond.### Solving the System of EquationsCombining these, you derive a system of equations, and solving them provides the minimum initial velocity to clear the fence. This involves solving a quadratic equation, often derived from adjustments for time in both equations.Once the initial velocity is known, you use it to solve for the total distance, especially focusing on where the rock will land after passing over the fence. This involves calculating additional time of flight when the rock falls from its highest point back to ground level:\[0 = h_f + v_0 \sin(\theta)t' - \frac{1}{2}gt'^2 \]These equations are critical for accurately predicting motion in two dimensions during projectile scenarios.