Question

1. a skydiver at rest jumps out of an airplane at 4,000 meters. how fast will they be traveling in meters per second after 4 seconds?

○ −39.24 m/s
○ −29.43 m/s
○ −19.62 m/s
○ −49.05 m/s

Explanation:

Step1: Identify the formula for free - fall velocity

The skydiver is in free - fall (assuming no air resistance initially). The formula for the velocity of an object in free - fall is \(v = u+at\), where \(u\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(t\) is the time. The skydiver starts from rest, so \(u = 0\ m/s\). The acceleration due to gravity \(g=- 9.81\ m/s^{2}\) (negative because it is in the downward direction, assuming upward as positive). The time \(t = 4\ s\).

Step2: Substitute the values into the formula

Substitute \(u = 0\), \(a=-9.81\ m/s^{2}\) and \(t = 4\ s\) into the formula \(v=u + at\).
\(v=0+(-9.81)\times4\)
\(v=-9.81\times4=- 39.24\ m/s\)

Answer:

\(-39.24\ m/s\) (the option corresponding to \(-39.24\ m/s\))