By equation of motion we have S = ut + 0.5 a[tex]t^{2}[/tex]
Where u = Initail velocity, S= dispalcement, t = time taken, a = acceleration
Here S = 3500-400 = 3100m, u = 0m/s, g = 9.8 m/[tex]s^{2}[/tex]
So, 3100 = 0+0.5*9.8*[tex]t^{2}[/tex]
[tex]t^{2}[/tex] = 3100/4.9
t = [tex]\sqrt{3100/4.9}[/tex]
t = 25.15 seconds
The time is taken by the skydiver to fall from a plane at 3500 m to an altitude of 400 m, where she will pull her ripcord is 25.139 sec.
Given to us
Height of the plane, S = 3,500 m,
The altitude where she will pull her ripcord, s = 400m
We can solve the problem using the second equation of motion, according to the second equation of motion,
[tex]S=ut+\dfrac{1}{2}at^2[/tex]
Substitute the value,
[tex]S-s=ut+\dfrac{1}{2}gt^2[/tex]
[tex]3,500-400=(0)t+\dfrac{1}{2}(9.81)t^2[/tex]
t = 25.139 sec.
Hence, the time is taken by the skydiver to fall from a plane at 3500 m to an altitude of 400 m, where she will pull her ripcord is 25.139 sec.
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