Awesome! The ball returns to its initial position. So, both the initial and final positions are \(y=10\). Therefore, the displacement is \(10-10=0\).

We talked about speed earlier. Let's introduce a related concept called

So, to calculate velocity, calculate the displacement and divide it by the time taken. \(\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}\). In calculus terms, \(v = \frac{dr}{dt}\), where \(v\) is the velocity, \(r\) is the position and \(t\) is the time.

**velocity**. You will use velocity throughout Physics. So, it is important that you understand its meaning thoroughly! Velocity is just the**rate of change of displacement**.So, to calculate velocity, calculate the displacement and divide it by the time taken. \(\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}\). In calculus terms, \(v = \frac{dr}{dt}\), where \(v\) is the velocity, \(r\) is the position and \(t\) is the time.

If the ball takes \(5 \, \mathrm{s}\) to return, which is correct?

\(\mathrm{Velocity} \, = 4 \, \mathrm{m/s}\), \(\mathrm{Speed} \, = 4 \mathrm{m/s}\)

\(\mathrm{Velocity} \, = 0 \, \mathrm{m/s}\), \(\mathrm{Speed} \, = 4 \mathrm{m/s}\)

\(\mathrm{Velocity} \, = 4 \, \mathrm{m/s}\), \(\mathrm{Speed} \, = 0 \mathrm{m/s}\)

\(\mathrm{Velocity} \, = 0 \, \mathrm{m/s}\), \(\mathrm{Speed} \, = 0 \mathrm{m/s}\)