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Question 22

A deep-sea research probe is released from rest at the ocean surface and descends vertically. The probe is designed such that it experiences a constant downward acceleration of 3.2 m s−23.2\text{ m s}^{-2}3.2 m s−2 under the combined effects of gravity, buoyancy, and drag.

Calculate the final speed of the probe when it has descended a distance of 45 m45\text{ m}45 m. Give your answer to 2 significant figures.

Use the equation:

v2−u2=2as v^2 - u^2 = 2as v2−u2=2as

where:

  • vvv is the final speed
  • uuu is the initial speed
  • aaa is the acceleration
  • sss is the distance
[3]
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