Question

A body of mass 2 kg is dropped from height of 22.5 metres. It hits the ground in 3 s. If g = 10 m/s² , what is the force that air resistance offers to this object?
(Assume constant air resistance)

Answer 1

Given:

Mass of the body, M = 2 kg

Initial velocity of the body, u = 0

Height from which the ball is dropped, H = 22.5 m

Let the acceleration of the body be a (in downward direction).

Calculating acceleration of the body (assuming constant)

H = ut + $\frac{1}{2}$ at²

22.5 = $\frac{1}{2}$a (3)²

=> a = 5 m/s²

Assuming the air resistance to be constant.

Let the air resistance be R.

There are two forces on the body:

1. Weight in downward direction.

2. Force by air resistance in upward direction.

The force equation for the body can be written as:

Mg − R = Ma

$$\begin{gathered} \Rightarrow a=\frac{(Mg-R)}{M} \\ \Rightarrow 5=\frac{(20-R)}{2} \\ \Rightarrow 10=20-R \\ \Rightarrow R=10\mathrm{N} \end{gathered}$$

Force offered by the air resistance is 10 N.