A certain freely falling object, released from rest, requires 1.85 s to travel the last 26.5 m before it hits the ground. (a) Find the velocity of the object when it is 26.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) -2.70 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the total distance the object travels during the fall.

  a) -5.26 m/s

  b) 27.91 m

Explanation:

a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.

  average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s

The change in velocity in the first half of the interval is ...

  Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s

So, the initial velocity (at the beginning of the last 1.85 s interval) is ...

  v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)

  v1 = -5.259 m/s

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b) The velocity when the object hits the ground is ...

  v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s

This is related to the distance traveled by ...

  v² = 2dg . . . . . where g is the acceleration and d is the distance traveled

  d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters

The object travels a total distance of about 27.911 meters.

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The attached graph shows height vs. time.

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