Acceleration Due to Gravity

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Uniform Acceleration Due to Gravity in Free Fall

Galileo was the first to measure how objects fell to Earth. He discovered that an object, regardless of its mass, fell downward a distance that is proportional to the square of the time elapsed. In other words, an object's velocity uniformly increases as a function of time called acceleration (a). Note: An object, such as a feather, would be adversely affected by air resistance and would not be considered a body in free fall.

We call the acceleration of a body in free fall, the uniform acceleration due to gravity (g). However, the value for the gravitational acceleration (g) is not constant and is dependent on where you are measuring an object's acceleration.

g (on the Moon) = 1.6 m/s2

g (on the Earth) = 9.8 m/s2 = 32 ft/s2

Note: The uniform acceleration due to gravity (g) is expressed as an average value. The amount of mass under your feet, such as standing on top of a mountain vs. at the shore, will cause (g) on Earth to vary between 9.787 and 9.808 m/s2.

Example 1.

Galileo drops, not throughs, a cannonball off of a balcony and it takes 3 seconds to impact the ground. How far did the cannonball fall?

Note: A graph often represents the relationship between two or more things. In this case, vertical motion (up and down) is plotted along the y-axis, and horizontal motion (forward and backward) is plotted along the x-axis of the graph. We will adopt, for this example, the convention that down is positive and up is negative. Therefore the acceleration due to gravity (g), toward the center of the Earth, is also positive.

We know the initial velocity of the cannonball is v0 = 0, the cannonball's acceleration due to gravity on Earth is a = g = 9.8m/s2, the time the cannonball is in free fall is t = 3.0 s, and the distance the cannonball falls along the y-axis is y =? We can find y with the equation:

y = v0 + ½at2

y = 0 + ½ (9.8m/s2)(3.0s)2

Note: Seconds (s2) cancel out leaving meters (m).

We find that: y = (½)(9.8m)(9.0) = 44.1m

Note: The mass of the object is not a factor in the equation for universal acceleration due to gravity and a 10 kg cannonball falls at the same rate as a 1 kg ball.

Example 2.

A soccer ball is tossed straight up into the air at a velocity of 20m/s. How high does it go?

Point A is the starting or initial point where the soccer ball is tossed up into the air, point B represents the highest point the ball reaches where its velocity (v) = 0, and point C is the point where the soccer player catches the falling ball. Note: For this example, we will designate up as positive, and the initial velocity of the displacement vector will also be positive. The acceleration due to gravity is downward and will be negative.

We know the initial velocity is v0 = 20m/s, the final velocity at the highest point is vf = 0, and the acceleration due to gravity on Earth is a = g = -9.8m/s2.

vf2 - v02 = 2ay

0 - 400m2/s2 = 2(-9.8m/s2)y

Note: Seconds (s2) cancel out and (m2) becomes meters (m).

We find that: y = (-400m2/s2) / (-19.6m/s2) = 20.4m

Note: The acceleration due to gravity is the same for the soccer ball when it is tossed up (A) and then falls down toward the ground and caught (C). Therefore the velocities of the ball at point A and point C are the same but in opposite directions.

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