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Answer:
2560m or 2.56km (rounded to 3 significant figures)
Explanation:
First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):
[tex]Vertical \ velocity \ (\frac{m}{s} ) = u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) = u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) = a_{v} = 9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) = a_{h} = 0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}[/tex]
The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;
First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;
Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);
What we have is the horizontal velocity but we don't have the time taken;
One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;
What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:
800 = (0)t + ¹/₂·(9.8)t²
800 = 4.9t²
t² = 163.26...
t = 12.77...
We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:
[tex]u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...[/tex]