The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
<h3>
Time of motion of the girl</h3>
The time of motion of the girl is calculated as follows;
h = vt + ¹/₂gt²
where;
- v is initial vertical velocity = 0
- t is time of motion
- g is acceleration due to gravity
Substitute the given parameters and solve for time of motion;
50.8 = 0 + ¹/₂(9.8)t²
2(50.8) = 9.8t²
101.6 = 9.8t²
t² = 101.6/9.8
t² = 10.367
t = √10.367
t = 3.22 seconds
<h3>Final vertical velocity of the skydiver</h3>
vf = vi + gt
where;
vi is the initial vertical velocity = 0
vf = 0 + 9.8(3.22)
vf = 31.56 m/s
Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
Learn more about vertical velocity here: brainly.com/question/24949996
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I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
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