91Ó°ÊÓ

1. The time delay for spectator A is$\mathrm{Δ}{t}_A=0.23\text{s}$.
2. The time delay for spectator B is$\mathrm{Δ}{t}_B=0.12\text{s}$.
3. The slight line from the two spectators to the player kicking the ball meet at an angle of 90⁰ .

The time for sound to travel from the kicker to a spectator is given by,

$t_{sound}=\frac{d}{v}$

The time required for light to travel the same distance dis given by,

$t_{light}=\frac{d}{c}$

Here, c is the speed of light, d is the distance and v is speed of sound.

The delay between seeing and hearing the kick is,

$$\begin{gathered} \mathrm{Δ}t=t_{sound}-t_{light} \\ \mathrm{Δ}t=\frac{d}{v}-\frac{d}{c} \end{gathered}$$

But, v<<c, so delay can be approximated by,

$$\begin{gathered} \mathrm{Δ}t=\frac{d}{v} \\ d=v\mathrm{Δ}t \end{gathered}$$

Thus, the distance from the kicker to spectator A is,

$d_A=v\mathrm{Δ}{t}_A$

Substitute 343 m/s for v and 0.23 s for $∆t_a$into the above equation

$$\begin{gathered} d_A=343×0.23 \\ =79\text{m} \end{gathered}$$

Hence, spectator A is 79m far from the player.

Similarly, the distance from the kicker to spectator is,

$d_B=v\mathrm{Δ}{t}_B$

Substitute 343 m/s for v and 0. 12s for $∆t_b$into the above equation

$$\begin{gathered} d_B=343×0.12 \\ =41 \text{m} \end{gathered}$$

Hence, spectator B is 41m far from the player.

The line from the kicker to each spectator and from one spectator to the other, form a right angle with the line joining the spectators as the hypotenuse. So, the distance between the spectators is,

$D=\sqrt{{d_A}^2+{d_B}^2}$

Substitute 79 m for d_A and 41 for d_b into the above equation,

$$\begin{gathered} D=\sqrt{{79}^2+{41}^2} \\ =89\text{m} \end{gathered}$$

Hence, the spectators are far from each other.