A skydiver jumps out of a helicopter and falls freely for 5.6 s before opening the parachute. (a) What is the skydiver's downward velocity when the parachute opens? Correct: Your answer is correct. m/s (b) How far below the helicopter is the skydiver when the parachute opens? Incorrect: Your answer is incorrect. m

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;

The distance traveled by the skydiver during this time is calculated as follows;

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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.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)