Askydiver 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?

The acceleration of the sky diver, five seconds after jumping, is 4.8 m/s².

The given parameters;

The net force on the skydiver is calculated as follows;

Fₙ = W - F

Fₙ = mg - F

Fₙ = (60 x 9.8) - 300

Fₙ = 288 N

The acceleration of the sky diver, five seconds after jumping, is calculated as follows;

F = ma

Thus, the acceleration of the sky diver, five seconds after jumping, is 4.8 m/s².

Learn more here:link

-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.

they should be similar or the same

no sign of dia of fuel cell