IB MYP 4-5 Physics- Free fall motion- Study Notes - New Syllabus
IB MYP 4-5 Physics- Free fall motion- Study Notes
Key Concepts
- Free Fall Motion
Free Fall Motion
Free Fall Motion
Free fall is the motion of an object under the influence of gravity only, with no air resistance. The object experiences constant downward acceleration due to gravity, denoted as \( g \).
Key Characteristics:
- Acceleration \( a = g = 9.8\,\text{m/s}^2 \) (downward)
- Objects fall vertically downward or are projected vertically upward
- Upward motion: \( a = -g \) (retardation)
- At the topmost point, \( v = 0 \)
- Free fall is a type of uniformly accelerated motion
Equations of Motion for Free Fall:
- \( v = u + gt \)
- \( s = ut + \dfrac{1}{2}gt^2 \)
- \( v^2 = u^2 + 2gs \)
Take upward direction as negative if object is thrown upward.
Example:
A stone is dropped from a height of \( 20\,\text{m} \). How long does it take to reach the ground?
▶️ Answer/Explanation
Given: \( u = 0\,\text{m/s}, s = 20\,\text{m}, g = 9.8\,\text{m/s}^2 \)
Use \( s = ut + \dfrac{1}{2}gt^2 \):
\( 20 = 0 + \dfrac{1}{2} \cdot 9.8 \cdot t^2 \Rightarrow t^2 = \dfrac{40}{9.8} \approx 4.08 \)
\( t = \sqrt{4.08} \approx \boxed{2.02\,\text{s}} \)
Example:
A ball is thrown vertically upward with speed \( 20\,\text{m/s} \). How high does it rise?
▶️ Answer/Explanation
At the highest point, \( v = 0 \). Given: \( u = 20\,\text{m/s}, a = -9.8\,\text{m/s}^2 \)
Use \( v^2 = u^2 + 2as \):
\( 0 = 20^2 + 2(-9.8)s \Rightarrow -400 = -19.6s \Rightarrow s = \dfrac{400}{19.6} \approx \boxed{20.4\,\text{m}} \)
Example:
A stone is thrown vertically upward with speed \( 15\,\text{m/s} \) from the top of a tower \( 20\,\text{m} \) high. How long does it take to reach the ground?
▶️ Answer/Explanation
Step 1: Use sign convention
Upward is taken as positive. So:
\( u = +15\,\text{m/s},\quad g = -9.8\,\text{m/s}^2,\quad s = -20\,\text{m} \) (displacement to ground is downward)
Step 2: Use equation \( s = ut + \dfrac{1}{2}gt^2 \)
\( -20 = 15t + \dfrac{1}{2}(-9.8)t^2 \Rightarrow -20 = 15t – 4.9t^2 \)
Step 3: Rearranging:
\( 4.9t^2 – 15t – 20 = 0 \)
Step 4: Solve using quadratic formula
\( t = \dfrac{15 \pm \sqrt{(-15)^2 – 4(4.9)(-20)}}{2 \cdot 4.9} = \dfrac{15 \pm \sqrt{225 + 392}}{9.8} \)
\( = \dfrac{15 \pm \sqrt{617}}{9.8} \approx \dfrac{15 \pm 24.82}{9.8} \)
Two roots:
- Negative root: \( t \approx \dfrac{15 – 24.82}{9.8} \approx -1.0\,\text{s} \) → Discard
- Valid root: \( t \approx \dfrac{15 + 24.82}{9.8} \approx \dfrac{39.82}{9.8} \approx \boxed{4.06\,\text{s}} \)
Free Fall with Air Resistance
In real situations, falling objects experience air resistance (a type of drag force) that acts upward, opposite to the direction of motion. This force increases with speed and affects the motion significantly.
- Without air resistance, the object accelerates uniformly at \( g = 9.8\,\text{m/s}^2 \).
- With air resistance, the acceleration decreases over time as drag increases.
- Eventually, the object stops accelerating and falls at a constant speed. This is called terminal velocity.
Forces Acting During Free Fall with Air Resistance
- Weight: \( W = mg \) (acts downward)
- Air resistance: \( F_R \propto v^2 \) (acts upward and increases with speed)
- Net Force: \( F_{\text{net}} = mg – F_R \)
Terminal Velocity
Terminal velocity is the maximum constant speed an object reaches when the force of air resistance equals the weight of the object, so net force and acceleration become zero.
When: \( F_R = mg \Rightarrow F_{\text{net}} = 0 \Rightarrow a = 0 \)
Then: The object moves at a constant speed \( v_t \), called terminal velocity.
Velocity–Time Graph (With Air Resistance)
- Initially, object accelerates rapidly (high \( a \)).
- As speed increases, air resistance increases, reducing acceleration.
- Eventually, acceleration becomes zero, and the graph levels off at terminal velocity.
Example: Object Falling with Air Resistance
Example:
A skydiver of mass \( 70\,\text{kg} \) jumps from a plane. Air resistance increases with speed and eventually balances the skydiver’s weight. Calculate the terminal velocity if the resistive force at terminal speed is equal to weight.
▶️ Answer/Explanation
At terminal velocity, \( F_R = mg \)
\( F_R = 70 \times 9.8 = 686\,\text{N} \)
So, terminal velocity is reached when the air resistance equals \( \boxed{686\,\text{N}} \)
(Exact value of terminal velocity depends on body area, drag coefficient, and air density; this problem focuses on the concept.)