Difficulty: Intermediate | Prerequisites: 1D kinematics, vector components, trigonometry
Projectile motion and free fall are the first real applications of the kinematics toolkit. Free fall is just constant-acceleration motion with a = g (downward), and projectile motion splits the problem into independent horizontal (constant velocity) and vertical (free-fall) parts. This section accounts for a large share of midterm questions. If you can handle the 1D kinematic equations confidently, projectile problems are the same equations applied twice, once for each direction.
In free fall, the only acceleration is g = 9.81 m/s² downward. For projectile motion, horizontal velocity is constant and vertical motion follows free-fall equations. The hang time of a projectile launched and caught at the same height depends only on the vertical component of velocity. Two objects dropped from the same height hit the ground at the same time regardless of horizontal speed, but the one with horizontal speed hits faster overall.
Free fall
Motion under the influence of gravity alone, with no air resistance. The acceleration is g = 9.81 m/s² directed downward, regardless of the object's mass or initial velocity.
Projectile motion
The motion of an object launched into the air that moves in two dimensions under gravity alone. Think of it as two independent problems stacked together: constant velocity horizontally and free fall vertically.
Hang time
The total time a projectile spends in the air. For a projectile launched and caught at the same height, hang time = 2v₀ sin θ / g.
Independence of horizontal and vertical motion
The principle that horizontal motion does not affect vertical motion and vice versa. In simple terms, a ball thrown sideways off a cliff hits the ground at the same time as one dropped straight down from the same height.
Splashdown speed (impact speed)
The magnitude of the velocity vector at the moment of landing. It is found from √(vₓ² + vᵧ²), combining both components at impact.
Q13: Time for a drop to fall 4.0 m from rest.
Use Δy = ½gt²: 4.0 = ½(9.81)t² → t² = 0.815 → t = 0.903 s.
Q14: Position and velocity of the second drop when the first hits the pool.
The drips are at a steady rate, so three drops are equally spaced in time. With drop 1 hitting and drop 3 just detaching, the time between drops is t/2 = 0.452 s.
Drop 2 has been falling for 0.452 s: position = ½(9.81)(0.452²) = 1.0 m below the stalactite (so 3.0 m above the pool). Velocity = 9.81 × 0.452 = 4.43 m/s downward.
Q15: Drops per minute.
One drop every 0.452 s gives 60/0.452 ≈ 133 drops per minute.
Ball 1 is thrown upward with speed v₀ and reaches maximum height h (where v = 0).
At that instant, ball 2 is thrown upward with the same v₀.
Ball 1 now falls from rest at height h. Ball 2 rises from ground level with speed v₀.
By symmetry of the kinematic equations, they meet when ball 1 has fallen a quarter of h and ball 2 has risen to three-quarters of h. The crossing point is above h/2. Answer: B (above h/2).
Intuition: ball 2 is moving fast at first and covers more than half the distance h before ball 1 (starting from rest) catches up.
Q8: A football is in the air for 4.50 s, kicked at 63.0° and caught at the same level.
Hang time T = 2v₀ sin θ / g → v₀ = gT / (2 sin θ) = 9.81 × 4.50 / (2 sin 63°) = 44.145 / 1.782 = 24.8 m/s. Answer: B.
Q10: A ball thrown straight up returns in 2.75 s. A second ball is thrown at 40.0° and also returns in 2.75 s.
For the straight-up ball: v₀₁ = gT/2 = 9.81 × 2.75 / 2 = 13.49 m/s.
For the angled ball: same hang time means same vertical component of velocity, so v₀₂ sin 40° = 13.49, giving v₀₂ = 13.49 / sin 40° = 13.49 / 0.6428 = 20.99 ≈ 21.0 m/s. Answer: A.
Diver 1 drops straight down from height h: final speed = √(2gh).
Diver 2 runs off horizontally with speed v₀: at splashdown, vₓ = v₀ and vᵧ = √(2gh).
Diver 2's total speed = √(v₀² + 2gh), which is greater than √(2gh).
Answer: A (diver 2's splashdown speed is greater). The horizontal speed adds to the total without affecting the vertical component.
\Delta y = v_{0y}\,t + \tfrac{1}{2}g\,t^2 \qquad (\text{taking downward as positive for free fall})v_y = v_{0y} + g\,tv_y^2 = v_{0y}^2 + 2g\,\Delta yT_{\text{hang}} = \frac{2\,v_0 \sin\theta}{g} \qquad (\text{launched and caught at same height})v_{\text{impact}} = \sqrt{v_x^2 + v_y^2}Free-fall timing is how engineers calibrate drop towers for material testing and how geologists measure the depth of wells (drop a stone, time the splash). Projectile motion governs everything from sports (calculating the launch angle for maximum range) to ballistics and even the trajectory of water from a garden hose.
Students often think a heavier object falls faster. In free fall (no air resistance), all objects accelerate at the same rate regardless of mass.
A common error is assuming that an object thrown horizontally off a cliff takes longer to hit the ground than one dropped straight down. The vertical motion is identical; the horizontal speed is irrelevant to the fall time.
Students sometimes forget that at the peak of a projectile's trajectory, only the vertical velocity is zero. The horizontal velocity is still v₀ cos θ, and the object is still moving.
Confusing speed with velocity at splashdown. Speed is always positive (it is the magnitude of velocity). The diver with horizontal speed has a larger speed at impact, even though the vertical component is the same.
⚠️ The stalactite drip problem (Q13 to Q15) is a multi-part free-fall question that tests whether you can work with evenly spaced time intervals. This type of linked problem is worth 15 points.
⚠️ Hang-time questions (find initial speed given angle and total flight time) appear regularly. Memorise T = 2v₀ sin θ / g and be ready to solve for v₀.
⚠️ The "two divers" conceptual question (Q9) tests the independence of horizontal and vertical motion. Expect a similar qualitative comparison on the exam.
True or false: A ball thrown horizontally from a building takes longer to hit the ground than one dropped from the same height. (False: same fall time.)
Fill in the blank: At the top of a projectile's arc, the vertical velocity is ____ and the horizontal velocity is ____. (zero; v₀ cos θ)
True or false: Doubling the launch angle always doubles the range. (False: range depends on sin 2θ, which peaks at 45°.)
Fill in the blank: A freely falling object starting from rest falls ____ m in the first second. (4.9 m, since Δy = ½ × 9.81 × 1² ≈ 4.9)
True or false: Two projectiles with the same hang time must have the same vertical component of initial velocity. (True.)
Q: A stone is dropped from rest from a height of 20 m. How long does it take to hit the ground?
A: Δy = ½gt² → 20 = ½(9.81)t² → t = √(40/9.81) = 2.02 s.
Q: A ball is kicked at 30° above the horizontal with an initial speed of 20 m/s. What is its hang time?
A: T = 2v₀ sin θ / g = 2(20) sin 30° / 9.81 = 20/9.81 = 2.04 s.
Q: An object is thrown horizontally at 15 m/s from a 45 m cliff. What is its speed just before hitting the ground?
A: vᵧ = √(2 × 9.81 × 45) = 29.7 m/s. Total speed = √(15² + 29.7²) = √(225 + 882) = √1107 = 33.3 m/s.
Q: A ball is thrown straight up at 20 m/s. At what height does it have a speed of 10 m/s?
A: v² = v₀² − 2gΔy → 100 = 400 − 2(9.81)Δy → Δy = 300/19.62 = 15.3 m.
Projectile motion is a direct extension of the vector-addition and 1D-kinematics skills from the previous notes. It also sets up the forces unit: once you add air resistance or friction, the acceleration is no longer simply g, and you will need Newton's second law to find it. The conceptual reasoning about the cliff divers (Q9) foreshadows energy conservation, where the same result follows from kinetic and potential energy arguments.
projectile motion, free fall, hang time, time of flight, range, trajectory, independence of motion, horizontal and vertical components, splashdown speed, impact speed, cliff diver problem, stalactite drip, g = 9.81, Physics 114, UW, midterm 1