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) \(v=\frac{2 R \sin (2 \theta)}{t \sin (\theta)}\)

Step by step solution

01

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: \(t \cdot \frac{R}{v^2\sin(2\theta)} = \frac{1}{2v \sin(\theta)}\)

02

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: \(v = \sqrt[3]{\frac{2tR}{\sin(\theta)\sin(2\theta)}}\)

03

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.

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