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A rock is thrown off a cliff with a speed of 1.7 m/s downward. How far will it fall after 4.2 seconds have elapsed?

a) 35.3 meters
b) 17.7 meters
c) 76.9 meters
d) 7.77 meters

User Jhylands
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1 Answer

3 votes

Final answer:

Calculating the distance a rock falls involves using the kinematic equation d = v_i * t + (1/2) * a * t^2. For a rock thrown downward at 1.7 m/s from a cliff, after 4.2 seconds, the calculated distance is 93.3 meters, which does not match any of the given options.

Step-by-step explanation:

The question pertains to the calculation of the distance a rock falls when thrown off a cliff with an initial downward velocity. To find the distance fallen after a certain time when the rock is thrown downward, we can use the kinematic equation for uniformly accelerated motion:

d = v_i * t + (1/2) * a * t^2

Where:

  • d represents the distance fallen
  • v_i is the initial velocity (1.7 m/s downward)
  • t is the time elapsed (4.2 seconds)
  • a is the acceleration due to gravity (9.8 m/s^2)

By substituting in the given values:

d = 1.7 * 4.2 + (1/2) * 9.8 * 4.2^2

d = 7.14 + (1/2) * 9.8 * 17.64

d = 7.14 + 86.1624

d = 93.3024 meters

However, none of the given options (a) 35.3 meters, (b) 17.7 meters, (c) 76.9 meters, (d) 7.77 meters matches the calculated distance. It appears there has been a miscalculation or the given options are incorrect. The correct answer should be 93.3 meters, provided that air resistance is ignored and the acceleration due to gravity is 9.8 m/s^2.

User Ramratan Gupta
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