Answer:
The horizontal distance traveled by the skydiver is 510.8 m.
Explanation:
Given;
height of fall, h = 800 m
initial velocity of the airplane, u = 40 m/s
The time to fall to the ground is calculated as;
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2 \times 800}{9.81} }\\\\t = 12.77 \ s[/tex]
The horizontal distance or range of the motion is calculated as;
R = ut
R = 40 m/s x 12.77 s
R = 510.8 m
Therefore, the horizontal distance traveled by the skydiver is 510.8 m.
