A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits
1 answer:
Answer:
Speed, Vfx = 7.619 m/s
Explanation:
Vertical distance, Dx = 5.4m
Horizontal distance, Dy = 8m
Acceleration due to gravity, g = 9.8m/s²
Initial speed, Vix = 0m/s²
To find the speed, we would use the second equation of motion to find the time, t;
Dx = Vixt + ½gt²
Substituting into the equation, we have;
5.4 = 0(t) + ½(9.8)*t²
5.4 = 0 + 4.9t²
Rearranging the equation, we have;
4.9t² = 5.4
t² = 5.4/4.9
t² = 1.1020
Taking the square root of both sides;
t = 1.050 secs.
For the speed;
Dy = Vfxt
Vfx = Dy/t
Vfx = 8/1.050
Vfx = 7.619 m/s
<em>Therefore, the speed of the pelican is 7.619 m/s</em>
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