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## UNSW Sydney Portrait of Galileo by Justus Sustermans

# Galileo's ball

Understanding what precisely happens to a ball if you toss it up in the air is crucial for physics. Qualitatively, we can say that the ball goes up, and then comes down. But can we be more precise? Galileo realized that it is a question of identifying the two quantities that describe the situation, and finding the mathematical relationship between them. And the genius of Newton came up with the physical laws that allowed the 17th century to fully resolve this essential question.

In this step we will

• 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?

Suppose that Galileo climbs to the top of the leaning tower of Pisa © “Tower of Pisa/Attribution-ShareAlike 2.0 Generic (CC BY-SA 2.0)

which has a maximum height of $$\normalsize{56\; m}$$, and leaning over the side, throws a ball up in the air so that it goes straight up, and then falls to the ground below. What actually happens to the position of the ball as a function of time? This is an important physics question.

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

Galileo realized that the mass of the ball does not affect how long it takes to get to the ground. Some years later, Isaac Newton described the exact nature of the gravitational force, and how force creates acceleration.

Newton’s laws showed that the really fundamental quantity was not position, as we might naively think, nor was it velocity. Rather it was the acceleration that was the key quantity, because it was determined by the forces acting on a particle.

Newton’s fundamental law of motion

$\Large{F=ma}$

for a mass $$\normalsize{m}$$ shows that if we know the force $$\normalsize{F}$$, then the acceleration $$\normalsize{a}$$ is also known. This is a very important example of a direct proportionality between acceleration and force, with the mass as the constant of proportionality. But this can also be viewed in another way.

For a particle near the surface of the earth, the acceleration is remarkably simple. Newton’s law of Gravitation showed that, on the surface of the earth, the force a particle feels is directly proportional to the mass, that is

$\Large{F_g=constant \times m}.$

When we put these two together, we deduce that the acceleration due to gravity, near the surface of the earth, is constant independent of mass.

It then followed, using a bit of calculus, that the dependence of velocity on time was necessarily a 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

To be specific, suppose that from a starting height of $$\normalsize{s=48}$$ metres, the ball had an initial upward velocity of $$\normalsize{8}$$ metres per seconds, or $$\normalsize{8\;m/s}$$.

From Newton’s reasoning, and some observation, we find that the constant acceleration due to gravity is (in units of m/sec^2)

$\Large{a=-10}.$

The velocity of Galileo’s ball, since it starts initially with a speed of $$\normalsize{8}$$, is then

$\Large{v=8-10t}$

and the position, which starts initially at $$\normalsize{s=48}$$, is then

$\Large{s=48+8t-5t^2}.$ This gives an explicit description of where the ball is at all times. Remarkably, the trajectory is independent of mass, and even more remarkably is completely determined by the relatively simple geometry of the parabola.

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?

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!

A2. Since

$\Large{48+8t-5t^2 = (4-t)(5t+12)}$

the $$\normalsize{t}$$-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.

A4. Following the same reasoning but with different numbers, we would have the velocity

$\Large{v=3-10t}$

and position

$\Large{s=20+3t-5t^2}.$