Freely Falling Bodies

1. Freely Falling Bodies In the absence of air resistance we find that all bodies, regardless of their size, weight or composition, fall with the same acceleration at the same point on the earth’s surface, and if the distance covered is not too great. The acceleration remains constant throughout the fall. This ideal motion, in which air resistance and the small change in acceleration with altitude is neglected, is called “free fall.” The acceleration of a freely falling body is called acceleration due to gravity and is denoted by g. Near the earth’s surface, its magnitude is approximately: g = 9.8 m/s2 = 980 cm/s2 = 32 ft/s2 and is directed down toward the center of the earth. As shown by the examples below, the equations of kinematics apply to freely falling bodies. Examples 1. A stone is dropped from rest from the top of a tall building. After 3 s of free fall, what is the displacement of the stone? Given:

v0 = 0 m/s; t = 3 s; ᾶ = -9.8 m/s2

Using (2.8) equation of kinematics, y = v0 t + ½ at2 = v0 + ½ gt2 = 0 + ½ (-9.8 m/s2)(3 s)2 = -44.1 m 2. After 3 s of free fall, what is the velocity of the stone in ex. 1? Using (2.4) equation of kinematics, v = v0 + at = v0 + gt v = 0 + (-9.8 m/s2)(3 s) = - 29.4 m/s.

The acceleration due to gravity is always a downward-pointing vector. It describes how the speed increases for an object that is falling freely. This same acceleration also describes how the speed decreases for an object moving upward under the influence of gravity alone, in which case the object comes to a momentary halt and then falls back to earth. 3. A man tosses a coin up with an initial speed of 5 m/s. In the absence of air resistance, how high does the coin go above its point of release? Using (2.9) equation of kinematics, v2 = vo2 + 2ay Solving for y, v2 – v02_ 2g = (0 m/s)2 - (5 m/s)2 = 1.28 m 2(-9.8 m/s2 ) 4. In ex. 3, how long is the coin in the air before returning to its release point? y=

Reasoning: During the time the coin travels upward, gravity causes its speed to decrease to zero. On the way down, however, gravity causes the coin to regain the lost speed. Thus, the time for the coin to go up is equal to the time for it to come down. In other words, the total travel time is twice the time for the upward motion. With the same data for the upward trip, we use eq. 2.4 (v = v0 + at) to find the upward travel time. Solution: Solving for t in eq. 2.4, t = (v – v0)/g = (0 m/s – 5 m/s)/(-9.8 m/s2) = 0.510 s

The total up-and-down time is twice this value or 1.02 s