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?

• 19.62 m/s
• 49.05 m/s
• 29.43 m/s
• 39.24 m/s

Explanation:

Step1: Identify the formula

Use the formula \(v = v_0+at\). Here, \(v_0 = 0\) m/s (starts from rest), \(a=- 9.81\) m/s² (acceleration due to gravity, negative as it is in the downward direction), and \(t = 4\) s.

Step2: Substitute values into the formula

Substitute \(v_0 = 0\), \(a=-9.81\) m/s², and \(t = 4\) s into \(v=v_0 + at\). We get \(v=0+(-9.81)\times4\).

Step3: Calculate the result

\(v=-9.81\times4=-39.24\) m/s.

Answer:

• \(39.24\) m/s