AP Physics Section 2-7 Falling Objects

Galileo’s experiments Galileo Galilei (1564–1642) performed many of the first experiments in physics, especially in relation to motion and force. Galileo did several experiments with falling objects and postulated a scientific law regarding the acceleration of falling objects. At a given location on the earth and in the absence of air resistance, all objects fall with the same constant acceleration. At a given location on the earth and in the absence of air resistance, all objects fall with the same constant acceleration.

Acceleration due to gravity Air creates a drag force on objects usually termed air resistance. Light objects with large surfaces areas are affected more by this. In the absence of air, such as on the moon, objects fall with constant acceleration. Air creates a drag force on objects usually termed air resistance. Light objects with large surfaces areas are affected more by this. In the absence of air, such as on the moon, objects fall with constant acceleration. This acceleration is the acceleration due to gravity. It is often given the symbol “g.” This acceleration is the acceleration due to gravity. It is often given the symbol “g.” This acceleration is the acceleration due to gravity. It is often given the symbol “g.” At the surface of the earth, the mean value is: g = 9.80 m/s2 (32 ft/s2) The value of g will vary with latitude, altitude, and with the nature of rock materials and metals underground. The value of g will vary with latitude, altitude, and with the nature of rock materials and metals underground. The value of g will vary with latitude, altitude, and with the nature of rock materials and metals underground.

Equations and coordinates We will make use of the four kinematic equations as well as the definition for constant acceleration that helped derive them. Since falling objects fall vertically, we often substitute y for x (and y0 for x0) in the equations. We can neglect air resistance, and assume that displacements, velocities and accelerations are only in the y direction. We will make use of the four kinematic equations as well as the definition for constant acceleration that helped derive them. Since falling objects fall vertically, we often substitute y for x (and y0 for x0) in the equations. We can neglect air resistance, and assume that displacements, velocities and accelerations are only in the y direction. The choice of the location of y0 is arbitrary, and chosen for convenience. It is often the ground. The choice of the location of y0 is arbitrary, and chosen for convenience. It is often the ground. The choice of the location of y0 is arbitrary, and chosen for convenience. It is often the ground. The positive direction may be up or down, but all vector values must be consistent with your choice. The positive direction may be up or down, but all vector values must be consistent with your choice.

Four Kinematic Equations The four kinematic in vertical form. variables y = y0 + ½ (v + v0)t 1 y, v, t v, a, t 2 v = v0 + at y = y0 + v0 t + ½at2 y, a, t 3 y, v, a 4 v2 = v0 2 + 2a(y – y0)

Dropping an object When dropping an object, the initial velocity in the y direction is zero. v0 = 0 m/s. v is often unknown. Acceleration = |g| = 9.80 m/s2. 1. How far will a metal ball fall in 3.50 s? Choose coordinate system: up is + , down is – Known values: Unknown value: v0 = 0 m/s a = –9.80 m/s2 y = ? t0 = 0 s t = 3.50 s y0 = 0 m y = (0.5)(–9.80 m/s2)(3.50 s)2 y = –60.0 m y = y0 + v0 t + ½at2 ie. 60 m below the drop point. y = ½at2

An object thrown downward In this case v0 is not equal to zero. 1. How far will a baseball fall in 2.30 s if thrown downward at a speed 20.0 m/s. Choose coordinate system: up is – , down is + Known values: v0 = +20.0 m/s Unknown value: t0 = 0 s a = +9.80 m/s2 y = ? y0 = 0 m t = 2.30 s y = y0 + v0 t + ½at2 y = v0 t + ½at2 y = (+20.0 m/s)(2.30s) + (0.5)(+9.80 m/s2)(2.30 s)2 y = (46.0 m) + (25.9 m) = +71.9 m ie. 71.9 m below the zero point.

2. What will the speed of the baseball be at this point? up is – , down is + Unknown value: v = ? Important: Since y was a calculated answer, it is better to use the original known values. v = v0 + at v0 = +20.0 m/s a = +9.80 m/s2 t = 2.30 s v = (+20.0 m/s) + (+9.80 m/s2)(2.30 s) v = (+20.0 m/s) + (+22.5 m/s) v = +42.5 m/s or a velocity of 42.5 m/s downward

An object thrown upward We will not consider forces, impulse, or motion during throwing, only motion after the object is tossed up. 1. A ball is thrown upward at 12.0 m/s. How high does it go (maximum height)? Choose coordinate system: up is + , down is – Known values: v0 = +12.0 m/s a = –9.80 m/s2 Unknown value: y = ? y0 = 0 m What can we at the assume about velocity at maximum height? v = 0.00 m/s v2 = v0 2 + 2a(y – y0) v2 – v0 2 = 2ay v2 – v0 2 – (+12.0 m/s)2 y = y = y = +7.35 m 2a 2(–9.80 m/s2)

2. How long is the ball in the air before it returns? (Total time of flight.) up is + , down is – Known values: v0 = +12.0 m/s a = –9.80 m/s2 y0 = 0 m y = 0 m Unknown value: t = ? y = y0 + v0 t + ½at2 Factor out a t and you’ll see that there are two times when y = 0. 0 = v0 t + ½at2 0 = t(v0 + ½at) t = 0s and v0 + ½at = 0 Ignoring the first possibility, we chose the second equation: v0 + ½at = 0 ½at = –v0 –2v0 a –2(12.0 m/s) t = = 2.45 s = –9.80 m/s2

3. How long does the ball take to reach max height? up is + , down is – Known values: v0 = +12.0 m/s a = –9.80 m/s2 y0 = 0 m v = 0.00 m/s Unknown value: t = ? v – v0 v = v0 + at t = a (0 m/s) – (12.0 m/s) –12.0 m/s t = = –9.80 m/s2 –9.80 m/s2 t = 1.22 s How does this time compare to the total time for the full flight? It’s half, but only if the full flight is back to the origin.

4. What is the velocity of the ball as it returns to the hand? up is + , down is – Known values: v0 = a = –9.80 m/s2 y0 = +12.0 m/s 0 m y = 0.00 m Unknown value: v = ? v2 = v0 2 + 2a(y – y0) v2 = v0 2 + 2a(y – y0) v2 = v0 2 |v| = |v0| Since the ball is now moving downward, v = –12.0 m/s It might be better to use the total time we found: 2.45 s v = v0 + at v = (+12.0 m/s) + (–9.80 m/s2)(2.45 s) v = –12.0 m/s or 12 m/s downward

Misconceptions Misconception 1: the acceleration and velocity vectors are always in the same direction. a and v are not necessarily in the same direction. When the ball is moving upward, the acceleration due to gravity points downward (created by the force of gravity downward). g always points downward! Misconception 2: the acceleration at the top of the flight is zero. Velocity decreases and reaches zero at the top of the flight. It then increases in speed as it moves downward. Both of these are due to constant downward acceleration. If a were equal to 0 at the top, the ball would stop and not fall back down!

displacement–time graph for vertical motion. Object goes up then down over time. velocity–time graph for vertical motion. Object decreases in velocity, stops, then increases velocity in the negative direction.

Using the quadratic formula 1. A rock is thrown vertically upward at 4.0 m/s from a 30. m cliff, and falls down to the base of the cliff. How long does this take? Choose coordinate system: up is + , down is – Known values: v0 = +4.0 m/s a = –9.80 m/s2 y0 = +30. m y = 0 m Unknown value: t = ? y = y0 + v0 t + ½at2 0 = 30 + 4 t + (–4.9)t2 This has quadratic form. c b a t = ⎧ ⎩ ⎨ Solve: –2.1 s t = –b ± √ b2 – 4ac 2a +2.9 s What does the negative value mean?

Graphs for a falling object slope of y–t graph gives the instantaneous velocity (derivative). slope of y–t graph gives the instantaneous velocity (derivative). slope of v–t graph gives the acceleration (derivative). Area under line gives displacement (integral). slope of v–t graph gives the acceleration (derivative). Area under line gives displacement (integral). slope of v–t graph gives the acceleration (derivative). Area under line gives displacement (integral). Area under the line on an a–t graph gives velocity for the time interval (integral). Area under the line on an a–t graph gives velocity for the time interval (integral).