An object falling straight down under the influence of gravity is an excellent example of constant acceleration in one dimension. Another is an object thrown straight up.

The acceleration comes from the gravitational force exerted on the object by the Earth. The magnitude is determined by the mass of the Earth, the radius of the Earth, and a number called the universal gravitational constant. At the Earth's surface, g, the acceleration due to gravity, equals 9.8 m/s² and is directed down.

Simultaneously drop a piece of paper and a textbook from the same height. Which hits the ground first?

1. The piece of paper
2. The textbook
3. Both hit at the same time

Why does the textbook hit the ground first?

1. Air resistance
2. It weighs more
3. Both 1 and 2
4. Some other reason

If the piece of paper is crumpled into a ball and then dropped with the book, which hits the ground first?

1. The piece of paper
2. The textbook
3. Both hit at the same time

You throw a ball straight up. It leaves your hand at 12.0 m/s.

(a) How high does it go?

(b) If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight?

(c) How fast is it traveling when you catch it?

Origin = height at which it leaves your hand

Positive direction = up

(a) How high does it go?

At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 and y_o = 0 into the equation:

v² = v_o² + 2 a (y - y_o)

This gives:

y = -v_o²/2a = -144/(-19.6) = 7.35 m.

The ball goes 7.35 m high.

(b) If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight?

Method 1: While going up, if the ball passes through a particular height at a particular velocity, on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. The down half of the trip is a mirror image of the up half. We could just figure out how long it takes to reach its maximum height, and then double that to get the total time.

Another method is to simply plug x = 0 into the equation:

x - x_o = v_o t + ½ a t²

This gives 0 = v_o t + ½ a t²

Cancelling a factor of t:

0 = v_o + ½ a t

t = -2v_o/a = -24/(-9.8) = 2.45 s.

(c) How fast is it traveling when you catch it?

The answer has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation:

v = v_o + a t

v = 12 - 9.8 (2.45) = -12 m/s.