A small fish is dropped by a pelican that is rising steadily at 0.500 m/s. How far below the pelican is the fish after 2.50 s?
1 answer:
Refer to the diagram shown below.
h = original height of the pelican when the fish is dropped (not relevant).
S = distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.
Use the following equation:
S = ut + (1/2)gt²
S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
= -29.375 m
The negative sign means that the fish drops by 29.375 m from the original height of h.
Answer: The fish is 29.375 m below where the pelican dropped it after 2.5 s.
h = original height of the pelican when the fish is dropped (not relevant).
S = distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.
Use the following equation:
S = ut + (1/2)gt²
S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
= -29.375 m
The negative sign means that the fish drops by 29.375 m from the original height of h.
Answer: The fish is 29.375 m below where the pelican dropped it after 2.5 s.
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