Grade 11 Physics – Unit 3.4
Vertical Motion (Free Fall)
👋 Hello grade 11! Today we’re talking about vertical motion – things falling or thrown upward under gravity. This is a very common topic in the entrance exam. Let’s make it simple and clear.
Freely Falling Bodies
Free fall means an object moves under the influence of gravity only (no air resistance). The acceleration is constant, directed downward, and is called acceleration due to gravity (g). Near Earth’s surface, \( g = 9.8 \, m/s^2 \) (sometimes they use 10 m/s² for simplicity).
Important: A freely falling object can be moving up, down, or just dropped. As long as only gravity acts, it’s free fall.
Equations for Vertical Motion
We use the same constant acceleration equations, but replace \( a \) with \( g \) (with sign). If we take upward as positive, then \( a = -g \).
\( v_f = v_i – g t \) (for upward positive)
\( y = v_i t – \frac{1}{2} g t^2 \) (displacement)
\( v_f^2 = v_i^2 – 2 g y \)
For an object thrown downward, both initial velocity and g are in the same direction, so we use +g : \( v_f = v_i + g t \), \( y = v_i t + \frac{1}{2} g t^2 \).
Example 3.8 – Ball thrown upward from a building
Situation: A particle is projected vertically upward with \( v_0 = 20 \, m/s \) from the top of a 15 m building (as in textbook figure). We want times to reach different levels.
Level A (starting point, y=0): Using \( y = v_i t – ½ g t^2 \), set y=0 → 0 = 20t – 5t² → t(20 – 5t)=0 → t=0 (start) or t=4 s (returns to starting point).
Level B (5 m above, y=+5): 5 = 20t – 5t² → divide 5: 1 = 4t – t² → t² -4t +1=0 → t = 2 ± √3 → t ≈ 0.27 s (going up) and t ≈ 3.73 s (coming down).
Level C (5 m below, y= –5): –5 = 20t – 5t² → divide 5: –1 = 4t – t² → t² -4t -1=0 → t = 2 ± √5 → t ≈ 4.24 s (only positive). So it takes 4.24 s to reach 5 m below.
Level D (ground, y= –15): –15 = 20t -5t² → 5t² -20t -15=0 → divide 5: t² -4t -3=0 → t = 2 ± √7 → t ≈ 4.65 s (positive).
Exercise 3.6 (from textbook)
A ball thrown upward at 10 m/s from a balcony 15 m high. Find (a) time to hit ground, (b) velocity on impact. (g=10)
Set y = -15 (downward displacement from balcony). Equation: \( -15 = 10t – 5t² \) → 5t² -10t -15=0 → t² -2t -3=0 → (t-3)(t+1)=0 → t=3 s (ignore -1). Velocity: \( v_f = 10 – 10×3 = -20 \, m/s \) (downward).
Graph of vertical throw (displacement & velocity)
Displacement-time: inverted parabola (symmetrical). Velocity-time: straight line with negative slope (–g). You already saw in section 3.3.
Activity 3.5 & 3.6 – measuring g & terminal velocity
Terminal velocity: When a body falls through air, drag force increases with speed. At some point, drag = weight, acceleration becomes zero, and it falls with constant speed called terminal velocity. For a person, terminal velocity is about 50 m/s; for a raindrop ~ 9 m/s. This explains why rain doesn’t hit like bullets.
Activity 3.6: Drop two balls of different mass (basketball & tennis). They hit at same time if air resistance is small – Galileo’s discovery. All objects fall at same rate in vacuum.
📘 In entrance exams, they may ask about terminal velocity: when acceleration = 0, forces balanced, velocity constant. They also like to ask about graphs for free fall with and without air resistance.
Review Questions 3.4 – vertical motion
Q2 (textbook): Ball thrown upward at 20 m/s from 25 m high building. (a) how high does it rise? (b) time to hit ground?
(a) From launch, additional height h = v²/(2g) = 400/20 = 20 m. So max height = 25+20 = 45 m above ground.
(b) To find time to ground: set y = -25 (displacement from launch point) → -25 = 20t -5t² → 5t² -20t -25=0 → t² -4t -5=0 → t = 5 s (positive).
Q3: Ball dropped from 10 m, rebounds to 2.5 m. Contact time 0.01 s. Find average acceleration during contact.
Velocity before impact: down \( v = \sqrt{2gh} = \sqrt{2×9.8×10} ≈ 14 \, m/s \). Velocity after rebound (up): \( v’ = \sqrt{2×9.8×2.5} ≈ 7 \, m/s \). Taking up as positive: Δv = 7 – (-14) = 21 m/s. a_avg = Δv/Δt = 21 / 0.01 = 2100 m/s² upward.
Activity 3.8 – Measuring reaction time using ruler drop
Reaction time \( t = \sqrt{\frac{2d}{g}} \). If you catch a ruler at d meters, that’s your reaction time. Reaction distance in driving = speed × reaction time. Very important for safety.
🔔 Exam tip: In “catch the ruler” problems, they give distance and ask for time, or give time and ask for distance. Just use \( d = \frac{1}{2} g t^2 \).
Summary of vertical motion
Always define a positive direction (usually up). g is downward, so with up positive a = –g. For objects thrown down, you may use g positive if you take down as positive – be consistent. Free fall equations are just special case of constant acceleration. Practice the sign convention.
✨ You now understand vertical motion! Next we’ll look at circular motion. But first, solve the exercises in your textbook – they are very similar to entrance exam questions. You’re doing great!