Question

Consider a projectile launched with velocity \(\left(v_{x 0}, v_{y_{0}}\right)\) from horizontal ground (with \(x\) measured horizontally and \(y\) vertically up). Assuming no air resistance, find how long the projectile is in the air and show that the distance it travels before landing (the horizontal range) is \(2 v_{x_{0}} v_{y_{0}} / g.\)

Short answer

The projectile is in the air for \(\frac{2v_{y0}}{g}\) seconds, and the horizontal range is \(\frac{2v_{x0}v_{y0}}{g}\).

Step-by-step solution

Understand the Variables

The given initial velocity is \(v_{x0}, v_{y0}\), where \(v_{x0}\) is the horizontal component and \(v_{y0}\) is the vertical component. The acceleration due to gravity is \(g\), acting downward with magnitude \(9.81 \, m/s^2\). The projectile is launched from the ground, so that the initial position is \(y = 0\).

Calculate Time of Flight

To find the time the projectile is in the air, use the vertical motion equation: \(y = v_{y0} t - \frac{1}{2} g t^2\). Set \(y = 0\) to solve for time, since the projectile starts and ends at ground level: \[ 0 = v_{y0} t - \frac{1}{2} g t^2 \]Factor the equation: \[ t(v_{y0} - \frac{1}{2} g t) = 0 \]This gives solutions \(t = 0\) (initial time) and \(t = \frac{2v_{y0}}{g}\). Thus, the time of flight is \(\frac{2v_{y0}}{g}\).

Determine the Horizontal Range

The horizontal range \(R\) is the horizontal distance the projectile travels during its flight.The horizontal motion is defined by \(x = v_{x0} t\), with \(v_{x0}\) constant because there's no horizontal acceleration.Substitute the time of flight (from Step 2): \[ R = v_{x0} \left(\frac{2v_{y0}}{g}\right) \]Simplify the expression to find the horizontal range: \[ R = \frac{2v_{x0} v_{y0}}{g} \]

Review the Results

We've found that the time of flight is \(\frac{2v_{y0}}{g}\) and the horizontal range is \(\frac{2v_{x0} v_{y0}}{g}\), as required by the problem statement. All calculations assume no air resistance and ideal projectile motion.

Key concepts

Time of Flight

The "time of flight" is a crucial concept in projectile motion, referring to the total time a projectile is in the air. To determine this, we focus on the motion in the vertical direction. The projectile is launched with an initial vertical velocity of \(v_{y0}\), and gravity acts downward with acceleration \(g = 9.81 \, \text{m/s}^2\). The vertical position of the projectile can be given by the equation:

• \(y = v_{y0} t - \frac{1}{2} g t^2\)

Since the projectile starts and ends at the ground level, we set \(y = 0\) to solve the equation for time. This gives us two solutions: one at the initial launch \(t = 0\), and the critical time just before it hits the ground again. Solving gives \(t = \frac{2v_{y0}}{g}\), which represents the total time the projectile remains airborne.
This concept of time of flight is important because it helps predict when and where the projectile will land, allowing for strategic planning in various applications such as sports or military operations.

Horizontal Range

The "horizontal range" of a projectile is the distance it travels along the horizontal axis before landing back on the ground. To find the horizontal range, we focus on the horizontal component of motion because it remains constant (no horizontal forces if air resistance is neglected).

• The horizontal motion is governed by: \(x = v_{x0} t\)

To find the full range of the projectile, we substitute the total time of flight from the vertical motion equation:

• \(R = v_{x0} \cdot \frac{2v_{y0}}{g}\)

This equation highlights how both the initial horizontal and vertical velocities contribute to how far the projectile will travel.
Understanding horizontal range is particularly beneficial in fields like engineering and gaming, where the accurate prediction of motion paths is crucial.

Kinematic Equations

Kinematic equations are vital tools in physics to analyze motion, especially when it comes to projectile motion. These equations allow us to relate time, velocity, and displacement through the consideration of acceleration and initial conditions. For any projectile, we typically use:

• \(y = v_{y0} t - \frac{1}{2} g t^2\)
• \(x = v_{x0} t\)

In the given exercise, these equations help us separate the motion into horizontal and vertical components, manage the displacement of the projectile, and solve for time of flight and horizontal range.
By mastering kinematic equations, students can predict the outcomes of varied projectile launched scenarios. This understanding is key in designing safe and efficient trajectories in disciplines ranging from sports to aerospace engineering.