Question

A ping pong ball (m = 0.0027 kg) is dropped and reaches the floor in 1.30 s. If, during its descent, the ping pong ball experiences an air resistance force of 0.022 N, calculate the displacement it fell. 

Answer 1

Let's apply the Newtons Second Law of Motion:

$$mg-f=ma,$$ $$a=g-\dfrac{f}{m}=9.8\ \dfrac{m}{s^2}-\dfrac{0.022\ N}{0.0027\ kg}=1.65\ \dfrac{m}{s^2}.$$

Then, we can find the vertical displacement:

$$y=\dfrac{1}{2}gt^2=\dfrac{1}{2}\cdot1.65\ \dfrac{m}{s^2}\cdot(1.3\ s)^2=1.4\ m.$$

Answer:

$y=1.4\ m.$