Chapter 8 Freefall and Projectile Motion

FREE FALL

Aristotle theory of motion
1. The fall of a heavy object toward the center of the earth is a natural motion because the object is just returning to its natural place.
2. Heavy objects fall faster than lighter ones. He explained that the increase in the rate of motion is proportional to the weight of the object.
3. Objects fall faster in air than in water. He assumed that the decrease in the rate motion is in proportion to the resisting force of the medium (air or water).
4. Objects sometimes move away from their natural force. He called this type of motion “ violent motion” which he explained was caused by a force.

Galileo theory of motion
1. The two objects (light and heavy) dropped simultaneously from the tower of Pisa fell and struck the earth at the same time.
2. When a ball was rolled down an inclined plane at fixed angle; the ratio of the distance covered to the square of the corresponding time was always the same.
3. When the angle of inclination is changed, the constant also changes but remains the same for the same angle.
4. The constant d/t2 is also the constant for falling object (refers to the acceleration due to gravity).

FREE FALLING BODY is uniformly accelerated because of gravity.

• If the motion of freely falling bodies is uniformly accelerated, then the formulas of free fall are the same as those of uniformly accelerated motion. Thus the equations are:

I. Vf = Vi + gt--------II. Vf = sq root (Vi^2 + 2gd)--------III. d = Vi t + 1/2 gt^2

• When the body starts from rest, Vi = 0
  

I. Vf = gt-------------II. Vf = sq root (2gd)-------------II III. d = 1/2 gt^2


PROBLEM SOLVING
1. A stone is dropped from a cliff 50 m high. Disregarding air resistance,
a. how long does it take the stone to reach the ground?
b. what is its speed when it strikes the ground?

Given: d = 50m and g = 10 m/s^2

Find a.) t and b.) Vf

For: a.)

Formula: d = ½ gt2

t = sq root (2d/g)----t t = sq root [2(50m)/10m/s^2]--------tt = 3.17 s

For: b.)
Vf = gt ----------------Vf= (10 m/s2)(3.17s) --------Vf= 31.7 m/s


2. A stone is thrown vertically downward with an initial velocity of 5 m/s. After 2 sec, a.) what is its velocity?, and b.) what distance did it cover?

Given: Vi = 5m/s t = 2s g = 10 m/s^2

Find Vf and d

Formula Vf = Vi + gt
Vf = 5m/s + 10 m/s^2 (2s)
Vf = 25 m/s

Formula: d = Vi t + ½ gt^2
d = (5m/s)(2s) + (10 m/s^2)(2s)^2
d = 30m