A skydiver with a mass of 60 kg jumps from an airplane. Five seconds after jumping

the force of air resistance is 300 N. What is the acceleration of the sky diver,

five seconds after jumping?

-4.8 m/s²

Explanation:

Apply Newton's second law:

∑F = ma

F - mg = ma

300 N - (60 kg) (9.8 m/s) = (60 kg) a

a = - 4.8 m/s²

The acceleration five seconds after jumping is 4.8 m/s² downward.

-4.8 m/s²

Explanation:

A skydiver with a mass of 60 kg jumps from an airplane. Five seconds after jumping the force of air resistance is 300 N. Therefore, the acceleration of the sky diver, five seconds after jumping is - 4.8 m/s². In order to find the answer, you must use Newton's second law. This is expressed as ∑F = ma.