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