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Whitepunk [10]
3 years ago
11

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

e time delay for spectator A is 0.201 s and for spectator B is 0.122 s. Sight lines from the two spectators to the player kicking the ball meet at an angle of 90°. How far are (a) spectator A and (b) spectator B from the player? (c) How far are the spectators from each other? (Take the speed of sound to be 343 m/s.)
Physics
1 answer:
NikAS [45]3 years ago
3 0

Answer:

a) 68.943 m

b) 41.846 m

c) 80.648 m

Explanation:

Given:

Delay in time for spectator A, t₁ = 0.201 s

Delay in time for spectator B, t₂ = 0.122 s

the delay in sound heard is the due to the distance being traveled by the sound from the kicker to the spectator

thus,

a) Distance of the kicker from A,

d₁ = speed of sound × time taken

d₁ = 343 m/s × 0.201 s = 68.943 m

b)  Distance of the kicker from B,

d₂ = speed of sound × time taken

d₂ = 343 m/s × 0.122 = 41.846 m

c) Since the angle between the two spectators for the player is 90°

thus, a right angles triangle is formed.

where, the distance between the spectators is the hypotenuses (s) of the so formed triangle

Therefore,

s² = d₁² + d₂²

on substituting the values, we get

s² = 68.943² + 41.846²

or

s² = 6504.22

or

s = √6504.22

or

s = 80.648 m

hence, the distance between the spectators is 80.648 m

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The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

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y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

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