Answer:
A. 54.88m/s.
B. 153.66m
Explanation:
Data obtained from the question include:
Time (t) = 5.6 secs
A. Determination of the skydiver's downward velocity.
Time (t) = 5.6 secs
Acceleration due to gravity (g) = 9.8m/s²
Velocity (v) =..?
v = gt
v = 9.8 x 5.6
v = 54.88m/s
Therefore, the skydiver's downward velocity is 54.88m/s
B. Determination of the height below the helicopter.
Time (t) = 5.6 secs
Acceleration due to gravity (g) = 9.8m/s²
Height (H) =..?
H = ½gt²
H = ½ x 9.8 x 5.6²
H = 153.66m
Therefore, the height below the helicopter when the parachute opens is 153.66m.
The final velocity of the skydiver when the parachute opens is 54.88 m/s.
The distance traveled by the skydiver during this time is 153.66 m.
The given parameters;
The final velocity of the skydiver when the parachute opens is calculated as follows;
[tex]v_f = v_0 + gt\\\\v_f = 0 + 9.8(5.6)\\\\v_f = 54.88 \ m/s[/tex]
The distance traveled by the skydiver during this time is calculated as follows;
[tex]h = v_0t + \frac{1}{2} gt^2\\\\h = 0 + (0.5 \times 9.8 \times 5.6^2)\\\\h = 153.66 \ m[/tex]
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