1. Kinematics IV
Free Fall motion

2. Falling Objects
In a vacuum (without air), all objects fall at the same rate.
Air Resistance accounts for the differences we see on earth.
For the most part, we will ignore air resistance by dealing with rigid bodies and slow speeds (Otherwise we need calculus!)

3. Every second, the velocity changes by 9.8m/s.
The acceleration due to gravity near the surface of the earth is:
𝒈=𝟗.𝟖𝒎/ 𝒔 𝟐
Every second, the velocity changes by 9.8m/s.

4. Drag (Air Resistance)
Note: This will not be on the test

5. Example: Falling from a tower
Problem: A ball is dropped from a tower m high. How far will the ball have fallen after a time t1 = 1.00s, t2 = 2.00s and t3 = s?

6. Approach 𝒗 𝒐 =𝟎𝒎/𝒔 Disregard air resistance
Acceleration is due to gravity, so 𝒂=−𝟗.𝟖𝒎/ 𝒔 𝟐

7. Solution 𝑦 𝑓 = 1 2 𝑎 𝑡 2 𝑦 1 = 1 2 ∗−9.80∗ 1.00 2 =−4.90𝑚
𝑦 2 = 1 2 ∗−9.80∗ =−19.6𝑚
𝑦 3 = 1 2 ∗−9.80∗ =−44.1𝑚

8. Example: Ball thrown upward
Problem: A person throws a ball upward into the air with an initial velocity of 15.0m/s. Find:
How high it goes
How long the ball is in the air before coming back down to his hand
How much time it takes for the ball to reach the maximum height
The final velocity when the ball reaches the hand

9. Approach Ignore air resistance
This time let’s use y-upward as positive
Assume a = -g = -9.80m/s
Use Kinematics EQ1, EQ2 and EQ3

10. Solution A) At the top, 𝒗=𝟎𝒎/𝒔, Using EQ3 and solving for y,
∆𝐲= 𝒗 𝟐 − 𝒗 𝒐 𝟐 𝟐𝒂 = 𝟎− 𝟏𝟓.𝟎 𝟐 𝟐 −𝟗.𝟖𝟎 =𝟏𝟏.𝟓𝐦
B) Since Displacement is 0 and using EQ2, we find
𝟎= 𝒗 𝒐 𝒕+ 𝟏 𝟐 𝒂 𝒕 𝟐 =𝒕 𝟏𝟓.𝟎−𝟒.𝟗𝟎𝒕 , 𝒕=𝟑.𝟎𝟔𝒔

11. C) Using EQ1 and knowing that 𝒗=𝟎𝒎/𝒔 at the top,
𝒕=− 𝒗 𝒐 𝒂 =− 𝟏𝟓.𝟎 −𝟗.𝟖𝟎 =𝟏.𝟓𝟑𝒔
NOTE: The time to reach the top is half the total time in air!
D) Using EQ1
𝒗= 𝒗 𝒐 +𝒂𝒕=𝟏𝟓.𝟎+ −𝟗.𝟖𝟎 𝟑.𝟎𝟔 =−𝟏𝟓.𝟎𝒎/𝒔
NOTE: The motion is parabolic. It comes back at the same speed it left with