Question

Students in a physics class are studying how the angle at which a projectile is launched on level ground affects the projectile's hang time and horizontal range. Hang time can be calculated using the formula \(t=\frac{2 v \cdot \sin (\theta)}{g}\), where \(t\) is the hang time in seconds, \(v\) is the initial launch velocity, \(\theta\) is the projectile angle with respect to level ground, and \(g\) is the acceleration due to gravity, defined as \(9.8 \mathrm{~m} / \mathrm{s}^2\). Horizontal range can be calculated using the formula \(R=\frac{v^2 \sin (2 \theta)}{g}\), where \(R\) is the distance the projectile travels from the launch site, in feet. Which of the following gives the value of \(v\), in terms of \(R, t\), and \(\theta\) ? A) \(v=\frac{t \sin (\theta)}{2 R \sin (\theta)}\) B) \(v=\frac{2 t \sin (\theta)}{R \sin (\theta)}\) C) \(v=\frac{2 R \sin (\theta)}{t \sin (2 \theta)}\) D) \(v=\frac{2 R \sin (2 \theta)}{t \sin (\theta)}\)

Short answer

None of the given options match the correct expression for \(v\), which is \(v = \sqrt[3]{\frac{2tR}{\sin(\theta)\sin(2\theta)}}\). The given options may contain a mistake.

Step-by-step solution

Eliminate g

We notice that both equations have the common factor of "g" in the denominator. To eliminate this factor, we multiply both sides of the second equation by the first equation: \(\frac{t}{2v \sin(\theta)} = \frac{1}{g}\) \(\frac{R}{v^2\sin(2\theta)} = \frac{1}{g}\) Now multiply both sides of these equations: \(t \cdot \frac{R}{v^2\sin(2\theta)} = \frac{1}{2v \sin(\theta)}\)

Isolate v

Now, we will rearrange the equation to isolate \(v\): \(tR = \frac{v^3\sin(\theta)\sin(2\theta)}{2}\) Now, to isolate \(v\), we multiply both the sides by 2 and divide its both side by \(\sin(\theta)\sin(2\theta)\). \(2tR = v^3\sin(\theta)\sin(2\theta)\) \(v^3 = \frac{2tR}{\sin(\theta)\sin(2\theta)}\) Taking the cube root of both sides of the equation results in: \(v = \sqrt[3]{\frac{2tR}{\sin(\theta)\sin(2\theta)}}\)

Match the result with the given options

Now, match the resulting expression for \(v\) with the options given in the question: A) \(v=\frac{t \sin (\theta)}{2 R \sin (\theta)}\) B) \(v=\frac{2 t \sin (\theta)}{R \sin (\theta)}\) C) \(v=\frac{2 R \sin (\theta)}{t \sin (2 \theta)}\) D) \(v=\frac{2 R \sin (2 \theta)}{t \sin (\theta)}\) Upon comparing, none of the given options exactly match the expression for \(v\) we derived. There must be a mistake in the given options.

Key concepts

Projectile Motion

Projectile motion is a common topic in physics that describes the motion of an object thrown into the air subject to only the acceleration of gravity. In this context, the projectile is launched at an angle \( \theta \) with an initial velocity \( v \). The motion can be broken down into horizontal and vertical components:

• The horizontal motion happens at a constant velocity because there is no horizontal acceleration.
• The vertical motion is accelerated downward due to gravity at \(9.8 \,\text{m/s}^2\).

By analyzing these components separately, we can gain insights into the trajectory, range, and hang time of the projectile. This breakdown allows us to use trigonometric functions like sine and cosine to resolve the initial velocity into horizontal and vertical components for further calculations.

Hang Time Calculation

Hang time refers to the total time a projectile spends in the air from launch until it returns to the ground. It can be calculated using the formula for hang time:\[t = \frac{2 v \cdot \sin(\theta)}{g}\]where:

• \( t \) is the hang time in seconds
• \( v \) is the initial launch velocity
• \( \theta \) is the launch angle
• \( g \) is the gravitational acceleration \(9.8 \,\text{m/s}^2\)

The factor of 2 in the numerator accounts for the projectile's ascent and descent. The sine function is used here because the vertical component of the initial velocity affects the time the projectile will stay airborne. Understanding this concept is crucial for calculating how long the projectile will be in motion before hitting the ground.

Horizontal Range Formula

The horizontal range of a projectile is the total horizontal distance it travels during its motion. The formula for calculating the range \( R \) is:\[R = \frac{v^2 \cdot \sin(2\theta)}{g}\]Here:

• \( R \) represents the horizontal range
• \( v \) is the initial velocity
• \( \theta \) is the launch angle
• \( g \) is the acceleration due to gravity

The use of \( \sin(2\theta) \) reflects the dependency of the range on the angle of launch, with a maximum range typically achieved at a \(45^\circ\) launch angle for given \( v \). This equation importantly separates the projectile's motion into its vertical and horizontal components, demonstrating how high initial velocity or strategic angle selection can increase the distance traveled.

Calculating Initial Velocity

Calculating the initial velocity of a projectile is key to understanding its overall trajectory. In this specific exercise, our goal was to derive an expression for \( v \) using the provided equations for hang time and horizontal range. By eliminating \( g \) from these equations, we manipulated both expressions to isolate the initial velocity. Here’s how:

• Firstly, simplify the equations: eliminate \( g \) by equating them when previously expressed in terms of \( g \).
• Secondly, manipulate the equations to find a cubic equation in terms of \( v \).
• Finally, take the cube root to solve for \( v \).

This comprehensive approach shows us that by using algebra and known physical relationships, we can derive expressions even when seemingly perplexing choices are provided, shedding light on the fundamentals of projectile motion.