# Galileo’s ball

- learn how to set up the appropriate time/position graph
- understand the basic logic that derives the quadratic dependence of position on time for a falling body from Newton’s laws of motion and gravitation.

## What goes up comes down

If we consider just a ball tossed up, moving in the \(\normalsize{y}\)-direction, then clearly the ball goes up, reaches a maximum, and then comes down. But this kind of qualitative description does not get us very far. How does it go up? How does it come down? Galileo Galilei (1564- 1642) was one of the first to consider carefully what was going on.One of the key ideas is that everything depends on time \(\normalsize{t}\) – this is the independent variable, so it plays the role usually assigned to \(\normalsize{x}\) in the Cartesian set-up, and is recorded horizontally. We can then ask how position \(\normalsize{s}\) depends on time, and how speed, or velocity \(\normalsize{v}\) depends on time, and how the acceleration \(\normalsize{a}\) depends on time? (In fact even the definitions of velocity and acceleration require some careful thought.)Q1(E-C): When you throw a ball directly up in the air, does it take longer going up, or coming down, or are these times equal (assuming that you catch it at the same level you threw it)? What is your guess? Could you do an experiment to check your guess?

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## Newton’s laws and the constant acceleration due to gravity

*acceleration*that was the key quantity, because it was determined by the forces acting on a particle.

*independent of mass*.

*linear function*, and that the dependence of position on time was necessarily a

*quadratic function*. These are relatively simple consequences of the constancy of the acceleration, and they give us complete understanding of the motion of the ball.

## Galileo’s ball

Q2(M): At what time does the ball hit the ground?Q3.(M) What is the maximum height reached by the ball?

Q4(E): What would be the position function if the ball started at a height of \(\normalsize{s=20}\) with an initial speed of \(\normalsize{v=3}\) m/sec?

## Answers

A1.It turns out that (neglecting air resistance) the time taken to go up exactly equals the time taken to go down. By the end of this step, you should be able to see that this fact is related to the symmetry of the parabola!\[\Large{48+8t-5t^2 = (4-t)(5t+12)}\]A2.Sincethe \(\normalsize{t}\)-intercepts of this parabola are when \(\normalsize{t=4}\) and \(\normalsize{t=-\frac{12}{5}}\). So the ball reaches the ground when \(\normalsize{t=4}\). Note that the solution \(\normalsize{t=-\frac{12}{5}}\) is not physically relevant, since the ball was tossed at \(\normalsize{t=0}\).A3.The maximum is achieved when \(t\) is exactly halfway between the two zeroes, namely \(\normalsize{t=(-\frac{12}{5}+4)/2=\frac{4}{5}}\), at which point the ball is at a position of \(\normalsize{\frac{256}{5} = 51.2}\) metres above the ground. Another very insightful way of getting this number is to observe that when the ball is at its maximum, the velocity \(\normalsize{v=8-10t}\) is instantaneously zero, which also gives \(\normalsize{t=\frac{8}{10}=\frac{4}{5}}\). Of course this requires knowing also the formula for the velocity.\[\Large{v=3-10t}\]A4.Following the same reasoning but with different numbers, we would have the velocityand position\[\Large{s=20+3t-5t^2}.\]

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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