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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Answer:
The answer is A. Which is true
Answer:
the energy from the sun travel to earth the answer is A .through the radiation
Answer:
v = ![\left[\begin{array}{c}0.66&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0.66%260%5Cend%7Barray%7D%5Cright%5D)
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:
The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:
