Answer:
23.4 m/s
Explanation:
f = actual frequency of the wave = 6.2 x 10⁹ Hz
[tex]f_{app}[/tex] = frequency observed as the ball approach the radar
[tex]f_{rec}[/tex] = frequency observed as the ball recede away from the radar
V = speed of light
[tex]v[/tex] = speed of ball
B = beat frequency = 969 Hz
frequency observed as the ball approach the radar is given as
[tex]f_{app}=\frac{f(V+v)}{V}[/tex] eq-1
frequency observed as the ball recede the radar is given as
[tex]f_{rec}=\frac{f(V-v)}{V}[/tex] eq-2
Beat frequency is given as
[tex]B = f_{app} - f_{rec}[/tex]
Using eq-2 and eq-1
[tex]B = \frac{f(V+v)}{V}- \frac{f(V-v)}{V}[/tex]
inserting the values
[tex]969 = \frac{(6.2\times 10^{9})((3\times 10^{8})+v)}{(3\times 10^{8})}- \frac{(6.2\times 10^{9})((3\times 10^{8})-v)}{(3\times 10^{8})}[/tex]
[tex]v[/tex] = 23.4 m/s
