Problem 56 - Physics for Scientists and Engineers with Modern Physics

Question

(III) Derive a formula for the horizontal range \(R,\) of a projectile when it lands at a height \(h\) above its initial point. (For \(h<0,\) it lands a distance \(-h\) below the starting point.) Assume it is projected at an angle \(\theta_{0}\) with initial speed \(v_{0}\) .

Step-by-Step Solution

Answer

The range formula is \(R = \frac{v_{0} \cos\theta_{0}}{g} \left( v_0 \sin\theta_0 + \sqrt{(v_0 \sin\theta_0)^2 + 2gh} \right)\).

Step 1: Identify the Equations of Motion

The horizontal and vertical components of the initial velocity are given as \(v_{x} = v_{0} \cos\theta_{0}\) and \(v_{y} = v_{0} \sin\theta_{0}\), respectively. The equations of motion for the projectile are \(x = v_{x} t\) for horizontal motion and \(y = v_{y} t - \frac{1}{2}gt^2\) for vertical motion, where \(g\) is the acceleration due to gravity.

Step 2: Express Time of Flight in Terms of Vertical Variables

Since the projectile lands at a height \(h\), set \(y = h\) in the vertical motion equation: \(h = v_{0} \sin\theta_{0} t - \frac{1}{2}gt^2\). Solve this quadratic equation for time \(t\) of flight resulting in: \(t = \frac{v_0 \sin\theta_0 \pm \sqrt{(v_0 \sin\theta_0)^2 + 2gh}}{g}\).

Step 3: Choose the Positive Root for Time

Only the positive time makes physical sense, thus use: \[t = \frac{v_0 \sin\theta_0 + \sqrt{(v_0 \sin\theta_0)^2 + 2gh}}{g}\].

Step 4: Substitute Time into Horizontal Motion Equation

Using the horizontal motion equation \(x = v_{0} \cos\theta_{0} \cdot t\), substitute \(t\) from Step 3, yielding: \[R = v_{0} \cos\theta_{0} \left( \frac{v_0 \sin\theta_0 + \sqrt{(v_0 \sin\theta_0)^2 + 2gh}}{g} \right)\].

Step 5: Simplify the Expression for Range

Simplify the equation from Step 4 to derive the formula for range: \[R = \frac{v_{0} \cos\theta_{0}}{g} \left( v_0 \sin\theta_0 + \sqrt{(v_0 \sin\theta_0)^2 + 2gh} \right)\].

Key Concepts

Horizontal Range

The horizontal range of a projectile refers to the total horizontal distance it travels before hitting the ground. When discussing this in the context of projectile motion, the range depends on both the horizontal and vertical components of the initial velocity, as well as the acceleration due to gravity. This is because the projectile follows a parabolic path influenced by its initial speed, angle of projection, and height difference between the launch and landing points.

For a projectile launched at an angle \( \theta_{0} \) with an initial speed \( v_{0} \), the equation to determine the horizontal range \( R \) includes both of these variables. The simplified equation provided earlier gives a generalized formula:

• \( R = \frac{v_{0} \cos\theta_{0}}{g} \left( v_0 \sin\theta_0 + \sqrt{(v_0 \sin\theta_0)^2 + 2gh} \right) \)

Where:
- \( g \) is the acceleration due to gravity.
- \( h \) is the height difference between the landing and launch points.
Considering these factors allows us to calculate how far horizontally the projectile will land based on its initial conditions.

Equations of Motion

In analyzing projectile motion, the equations of motion are pivotal. They allow us to break down the movement into components that are easier to manage: horizontal motion and vertical motion. Each has its own simple rules due to the difference in forces acting on the projectile.

**Horizontal Motion** reflects uniform motion, meaning there is no acceleration acting in the horizontal direction (ignoring air resistance). Thus the equation used is:

• \( x = v_{x} t \)

Where \( v_{x} = v_{0} \cos\theta_{0} \) is the horizontal component of the initial velocity.

**Vertical Motion** is affected by gravitational acceleration, \( g \). Here, the equation becomes:

• \( y = v_{y} t - \frac{1}{2}gt^2 \)

Where \( v_{y} = v_{0} \sin\theta_{0} \) is the vertical component of the initial velocity.

By solving these equations together, we can find the time of flight and range for the projectile, taking into account any additional height the projectile might have traveled to.

Trajectory

The trajectory is the path that the projectile follows as it moves under the influence of gravity. It typically takes the shape of a parabola. Understanding this path is key to predicting where the projectile will land.

A projectile's trajectory is primarily determined by its initial speed, angle of launch, and any height difference between the launch and landing points. The vertical component of velocity, influenced by gravity, causes the projectile to rise to a peak and then descend, while the horizontal component, assuming no air resistance, remains constant.

To analyze the trajectory:

• The initial angle, \( \theta_{0} \), determines the proportion of speed dedicated to vertical versus horizontal movement.
• The peak of the trajectory is reached when the vertical component of velocity is zero.
• As the projectile descends, the increased vertical speed caused by gravity brings it down to the landing point.

By considering these aspects, you can understand and predict the course of the projectile, allowing for precise calculation of where it will land relative to its starting point.