A skydiver steps out of a helicopter hovering thousands of metres above ground. After 3.5 s of free fall the skydiver pulls the rip cord. What is the skydiver’s displacement during the

3.5 s of free fall?

To find the skydiver's displacement during the 3.5 seconds of free fall, we need to calculate the distance fallen during this time.

During free fall, the distance fallen can be calculated using the equation:

d = (1/2) * g * t^2

where:
d = distance fallen
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

Plugging in the values, we have:

d = (1/2) * 9.8 * (3.5)^2

Simplifying,

d = 0.5 * 9.8 * 12.25

d = 57.05 meters

Therefore, the skydiver's displacement during the 3.5 seconds of free fall is approximately 57.05 meters.

To find the skydiver's displacement during the 3.5 seconds of free fall, we need to use the equation of motion for free fall. The equation we will use is:

displacement (d) = initial velocity (v0) * time (t) + (1/2) * acceleration (a) * time (t)^2

In this case, the skydiver is in free fall, meaning the only force acting on them is gravity. The acceleration due to gravity is approximately 9.8 m/s^2.

Since the skydiver steps out of a helicopter, we assume their initial velocity is 0 m/s.

Plugging these values into the equation, we get:

displacement (d) = 0 * 3.5 + (1/2) * 9.8 * (3.5)^2
displacement (d) = 0 + (1/2) * 9.8 * 12.25
displacement (d) = 0 + 4.9 * 12.25
displacement (d) = 4.9 * 12.25
displacement (d) = 59.92 meters (rounded to two decimal places)

Therefore, the skydiver's displacement during the 3.5 seconds of free fall is approximately 59.92 meters.

4.9 t^2