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Answers
Explanation:
The soldier was 17 sec in air.
Explanation:
Given that,
Distance = 19.6 m
Decelerate = 1 m/s
Speed = 4.6 m/s
The distance covers 19.6 m in time t with the acceleration due to gravity.
Using equation of motion
s = ut+\dfrac{1}{2}gt^2s=ut+
2
1
gt
2
19.6=0+\dfrac{1}{2}\times9.8\times t^219.6=0+
2
1
×9.8×t
2
t=\sqrt{\dfrac{19.6}{4.9}}t=
4.9
19.6
t = 2\ sect=2 sec
The velocity after 2 sec
Using equation of motion
v = u+gtv=u+gt
v = 0+9.8\times2v=0+9.8×2
v = 19.6\ m/sv=19.6 m/s
When the parachutist open the parachute then the initial velocity is 19.6 m/s.
Decelerates at 1 m/s and reach the ground with the 4.6 m/s.
So, the final velocity is 4.6 m/s
The time taken is given by
v = u-at'v=u−at
′
4.6=19.6-1\times t'4.6=19.6−1×t
′
t'=4.6-19.6t
′
=4.6−19.6
t'=15\ st
′
=15 s
Where t' is the time taken to reach the ground
The total time is
T =t+t'T=t+t
′
T = 2+15T=2+15
T= 17\ sT=17 s
Hence, The soldier was 17 sec in air.