9702_s22_qp_12 Question 6

 The water surface in a deep well is 78.0 m below the top of the well. A person at the top of the well drops a heavy stone down the well. 

Air resistance is negligible. The speed of sound in the air is 330 m s^–1. 

What is the time interval between the person dropping the stone and hearing it hitting the water? 

A 3.75 s 

B 3.99 s 

C 4.19 s 

D 4.22 s

ANSWER: D

The Physics Behind

• The time interval x between the person dropping the stone and hearing it hitting the water can be determined by calculating
• the time t₁ when the stone hits the water and produce sound
• the time t₂ when sound travels upwards and  reach the person's ear
• adding t₁ and t₂.
• For time t₁, the motion is freefall/at uniform acceleration so use the equation of motion s = ut₁  + 1/2 at₁²

78.0 = 0 + 1/2 (9.8) t₁² 

t₁² =15.92

t₁ =  3.99 s

• For time t₂, sound is a wave unaffected by gravity so use the constant speed formula v = d/t₂

t₂ = 78.0 / 330

t₂ = 0.236 s

• Therefore, answer is D.