Answer:
The speed of the baseball is approximately 19.855 m/s
Explanation:
From the question, we have;
The frequency of the microwave beam emitted by the speed gun, f = 2.41 × 10¹⁰ Hz
The change in the frequency of the returning wave, Δf = +3190 Hz higher
The Doppler shift for the microwave frequency emitted by the speed gun which is then reflected back to the gun by the moving baseball is given by 2 shifts as follows;
[tex]\dfrac{\Delta f}{f} = \dfrac{2 \cdot v_{baseball}}{c}[/tex]
[tex]\therefore{\Delta f}{} = \dfrac{2 \cdot v_{baseball}}{c} \times f[/tex]
Where;
Δf = The change in frequency observed, known as the beat frequency = 3190 Hz
[tex]v_{baseball}[/tex] = The speed of the baseball
c = The speed of light = 3.0 × 10⁸ m/s
f = The frequency of the microwave beam = 2.41 × 10¹⁰ Hz
By plugging in the values, we have;
[tex]\therefore{\Delta f} = 3190 \ Hz = \dfrac{2 \cdot v_{baseball}}{3.0 \times 10^8 \ m/s} \times 2.41 \times 10^{10} \ Hz[/tex]
[tex]v_{baseball} = \dfrac{3190 \ Hz \times 3.0 \times 10^8 \ m/s }{2.41 \times 10^{10} \ Hz \times 2} \approx 19.855 \ m/s[/tex]
The speed of the baseball, [tex]v_{baseball}[/tex] ≈ 19.855 m/s
