Question

A ball is shot from a compressed-air gun at twice its terminalspeed.
(For the following, assume the drag force depends onv²)

1. What is the magnitude of the ball's initial acceleration, as amultiple of , if it is shotstraight up?Express your answer using one significantfigure.
=
2. What is the magnitude of the ball's initial acceleration, as amultiple of , if it is shotstraight down? Express your answer using one significantfigure.
=

Answer 1

The magnitude of the drag force is given by

Fd = -1/2*rho*v^2*Cd*A.

Where rho is the density of air, v is the velocity, Cd is the dragcoefficient, and A is the cross-sectional area.

At the terminal velocity, vt, this will be equal and opposite tothe gravitational force

1/2*rho*vt^2*Cd*A = m*g

Now, because the drag force is proportional to the square of thevelocity, if we double the velocity the magnitude of the drag forcewill increase by afactor of 4.

Fd = 1/2*rho*(2*vt)^2*Cd*A
Fd = 1/2*rho*4*vt^2*Cd*A
Fd = 4*(1/2*rho*vt^2*Cd*A)
Fd = 4*m*g

So the drag force is 4 times the acceleration due to gravity,and

m*a = 4*m*g
a = 4*g

the acceleration is 4*g.

This drag force will always act in the direction opposite to themotion of the ball. When the ball is shot straight up gravity andthe drag force will beacting in the same direction. So

a = 4*g + g
a = 5*g.

When the ball is fired downwards the drag force will be acting inthe opposite direction to gravity.

a = 4*g - g
a = 3*g

So number 1 = 5 and number 2 = 3