mMCQ.

Navigation Menu

Step into mMCQ.

Launch mMCQ. diagnostic

Explore mMCQ.

MDCAT prepFree DiagnosticPricing & SubscribeSign in

Resources

Terms & Conditions

mMCQ.

© 2021 - 2025 mMCQ.All rights reserved.

WhatsApp
A Levels Physics (9702)•9702/12/M/J/22
Question 6 from 9702/12/M/J/22

Explanation

Total time = fall time + sound travel time

Steps:

  • Calculate fall time: tf=2dg=2×789.8≈3.99t_f = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \times 78}{9.8}} \approx 3.99tf​=g2d​​=9.82×78​​≈3.99 s.
  • Calculate sound time: ts=dv=78340≈0.23t_s = \frac{d}{v} = \frac{78}{340} \approx 0.23ts​=vd​=34078​≈0.23 s.
  • Add times: total t=3.99+0.23=4.22t = 3.99 + 0.23 = 4.22t=3.99+0.23=4.22 s.
  • Match to choices: 4.22 s is option D.

Why D is correct:

  • Uses free-fall equation d=12gt2d = \frac{1}{2} g t^2d=21​gt2 with g=9.8g = 9.8g=9.8 m/s² and sound speed formula t=dvt = \frac{d}{v}t=vd​.

Why the others are wrong:

  • A: Ignores sound time, underestimates total.
  • B: Fall time only with g=9.8g = 9.8g=9.8, misses sound.
  • C: Uses g=10g = 10g=10 m/s², slightly overestimates fall time.

Final answer: D

Topic: Equations of motion

Practice more A Levels Physics (9702) questions on mMCQ.me