Problem 56 - Physics for Scientists and Engineers with Modern Physics

Question

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 horizontal range formula is \(R = \frac{v_0 \cos(\theta_0) (v_0 \sin(\theta_0) + \sqrt{(v_0 \sin(\theta_0))^2 + 2gh})}{g} \).

Step 1: Write the equations of motion

We begin by writing the equations for the horizontal and vertical motions of the projectile. For horizontal motion: \[ x(t) = v_0 \cos(\theta_0) \cdot t \]For vertical motion: \[ y(t) = v_0 \sin(\theta_0) \cdot t - \frac{1}{2}gt^2 \]where \(g\) is the acceleration due to gravity.

Step 2: Solve for time using vertical motion

We need to find the time \(t\) when the projectile reaches height \(h\). We set the vertical position equation equal to \(h\): \[ h = v_0 \sin(\theta_0) \cdot t - \frac{1}{2}gt^2 \]This is a quadratic equation in \(t\):\[ \frac{1}{2}gt^2 - v_0 \sin(\theta_0)t + h = 0 \]

Step 3: Use the quadratic formula

Solve the quadratic equation for \(t\): \[ t = \frac{v_0 \sin(\theta_0) \pm \sqrt{(v_0 \sin(\theta_0))^2 + 2gh}}{g} \]The two solutions correspond to the times when the projectile is at the height \(h\). The only relevant solution is the positive one.

Step 4: Find the horizontal range

The horizontal range \(R\) is the horizontal distance traveled when the projectile is at height \(h\). Substitute the positive time from Step 3 into the horizontal equation: \[ R = v_0 \cos(\theta_0) \cdot \frac{v_0 \sin(\theta_0) + \sqrt{(v_0 \sin(\theta_0))^2 + 2gh}}{g} \]Simplify this expression to get the final formula.

Key Concepts

Understanding Horizontal Range

When you're thinking about projectile motion, one important aspect to grasp is the idea of horizontal range, often denoted as \( R \). This is simply the total distance a projectile travels along the horizontal axis, starting from the point of launch until it lands. This can be influenced by several factors:

• The initial speed \( v_0 \) at which the projectile is launched.
• The angle of launch \( \theta_0 \), which determines the initial direction of the projectile.
• The height \( h \) where it lands, whether higher, lower, or at the same point of origin.

The horizontal range is crucial in many real-world applications, such as sports, engineering, and even video game design, where predicting where an object lands can be essential. To calculate the range, you need to determine the time the projectile spends in the air (time of flight) and use it within the equation of motion for horizontal displacement. Once you have that, you can find how far it travels horizontally. Remember, though, while calculating this, air resistance is usually not considered in basic projectile problems.

Equations of Motion Explained

The equations of motion are at the heart of solving projectile motion problems. They break down the components of motion into horizontal and vertical parts because these two directions do not affect each other in simple scenarios.
For horizontal motion, the equation is straightforward:
\[x(t) = v_0 \cos(\theta_0) \cdot t\]
This equation shows that the horizontal position \( x \) at any time \( t \) is a product of the initial speed, the cosine of the launch angle, and the elapsed time. The term \( v_0 \cos(\theta_0) \) is the constant horizontal velocity because there's no horizontal acceleration in an ideal setting.
For vertical motion, it becomes a bit more complex because gravity acts on the projectile:
\[y(t) = v_0 \sin(\theta_0) \cdot t - \frac{1}{2} g t^2\]
This equation includes the effect of gravity, represented by \( g \), which is subtracted because it decelerates the projectile upwards and accelerates it downward.
Understanding these equations is crucial for breaking down projectile motion, as they allow you to solve for various unknowns, such as time of flight, maximum height, and range.

Role of Acceleration Due to Gravity

Gravity is a key player in any discussion of projectile motion. Represented by \( g \) and usually approximated as \( 9.8 \text{ m/s}^2 \) on Earth, it's the constant force that pulls objects towards the surface. It affects only the vertical component of a projectile's motion, causing objects to follow a parabolic path.

• In the vertical motion equation \( y(t) = v_0 \sin(\theta_0) \cdot t - \frac{1}{2}gt^2 \), the term \(-\frac{1}{2}gt^2\) causes the upward motion to slow down and eventually reverse as the object comes back towards the ground.
• Gravity doesn’t affect the horizontal motion, allowing it to remain constant as \( v_0 \cos(\theta_0) \).

Simply put, gravity dictates how high and how long a projectile stays in the air, while the initial launch conditions dictate how far it goes horizontally.
Thus, fully understanding the effects of gravity is essential to predict and calculate the path and landing point of any projectile successfully.