a parachutist jumps from an aeroplane.he has a mass of 80kg.initially,he 'free-falls',before his parachute opens.what is his initial acceleration

All free-falls on earth starts at g (9.8 m/s²) if initiated near the surface.

At high elevations, g is slightly reduced because of the greater distance from the centre of the earth.

Subsequently, as velocity is substantial, air-resistance plays an important rôle in reducing the acceleration until the terminal velocity is reached when the acceleration is zero.

Read about it at the following link, which includes interesting parachute jumps at the end of the article:
http://en.wikipedia.org/wiki/Free_fall

To find the initial acceleration of the parachutist during free fall, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = m * a).

During free fall, the only force acting on the parachutist is the force of gravity. According to Newton's second law, the force of gravity is equal to the parachutist's mass multiplied by the acceleration due to gravity (F = m * g), where "g" is approximately equal to 9.8 m/s² on the surface of the Earth.

So, the equation becomes:
m * a = m * g

Given that the parachutist's mass is 80 kg, we can substitute this value into the equation:
80 kg * a = 80 kg * 9.8 m/s²

Now, we can cancel out the mass (m) on both sides of the equation:
a = 9.8 m/s²

Therefore, the initial acceleration of the parachutist during free fall is 9.8 m/s², which is equal to the acceleration due to gravity.