5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A) What is the accelerat

Question

5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)
What is the acceleration of the shell right after it is released? (B) Find the maximum height above the ground
reached by the shell. (C) How long does it take for the shell to return to a height of 12.5m above the ground?

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bonexptip 4 years 2021-07-19T03:33:06+00:00 1 Answers 14 views 0

Answers ( )

  1. thongdat2
    0
    2021-07-19T03:34:22+00:00
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    Answer:

    (B) 13.9 m

    (C) 1.06 s

    Explanation:

    Given:

    v₀ = 5.2 m/s

    y₀ = 12.5 m

    (A) The acceleration in free fall is -9.8 m/s².

    (B) At maximum height, v = 0 m/s.

    v² = v₀² + 2aΔy

    (0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

    y = 13.9 m

    (C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

    v = at + v₀

    -5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

    t = 1.06 s

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