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Answer: he did travel 15 meters.
Explanation:
We have the data:
Acceleration = a = 1.2 m/s^2
Time lapes = 3 seconds
Initial speed = 3.2 m/s.
Then we start writing the acceleration:
a(t) = 1.2 m/s^2
now for the velocity, we integrate over time:
v(t) = (1.2 m/s^2)*t + v0
with v0 = 3.2 m/s
v(t) = (1.2 m/s^2)*t + 3.2 m/s
For the position, we integrate again.
p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0
Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m
Then the displacement at t = 3s will be equal to p(3s).
p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m
Answer:
The helicopter was 1103.63 meters high when the package was dropped.
Explanation:
We consider positive speed as a downward movement
y: height (m)
t: time (s)
v₀: initial speed (m/s)
Δy = v₀t + 
gt²
Δy= 15
×15 s + ×9.81
×(15 s)²
Δy= 1103.63 m
