DPST1021 Chap.2 Projectile Motion and Relative Velocity
Projectile Motion and Relative Velocity
Lecture 3 of Physics 1A takes the straight-line kinematics of the first week and lets it run in two or three dimensions. Position, velocity and acceleration become vectors, and the constant-acceleration equations from the formula sheet are applied to each axis separately.
A launched ball, a stone off a ledge or a parcel released from an aircraft all follow the same pattern: steady horizontal velocity, uniformly changing vertical velocity, and a single time variable tying the two together.
Projectiles fill most of the chapter.
You resolve a launch velocity into components, find a flight time from the vertical motion, carry that time across to the horizontal motion, and rebuild a final speed and direction with Pythagoras.
Level-ground results such as the maximum height and the time of flight follow quickly, but problems with unequal launch and landing heights need the full quadratic, and choosing the correct root is part of the skill.
Relative velocity completes the lecture.
Starting from the difference of two position vectors and differentiating gives the rule that the velocity of A relative to B equals the velocity of A minus the velocity of B. Rain seen from a moving car, an aircraft correcting for wind and a swimmer crossing a current all reduce to that single vector subtraction, done in components or with a triangle.
Everything here belongs to the mechanics block assessed in Test 1, and the first weekly homework quiz covers Lectures 1 to 3.
What this chapter covers
- 01
Vector kinematics: write position, velocity and acceleration as vectors in i, j and k, and apply each constant-acceleration equation to one axis at a time.
- 02
Launch components: resolve an initial speed and angle into horizontal and vertical parts using cosine and sine, keeping the sign of each part.
- 03
Independent axes: treat horizontal motion as constant velocity and vertical motion as free fall with g = 9.80 m/s squared, sharing only the elapsed time.
- 04
Unequal heights: solve the vertical quadratic for the flight time when launch and landing heights differ, and keep only the root that makes physical sense.
- 05
Impact velocity: rebuild the final speed and direction from the two components, and confirm the speed with an energy check where possible.
- 06
Relative velocity: derive the velocity of A relative to B from the difference of position vectors, then subtract velocities to solve wind, current and rain problems.
Worked example · free
A supply crate released from a level aircraft
- 1At release the crate shares the aircraft velocity, so vx0 = 60.0 m/s and vy0 = 0. Take the release point as origin with up positive.
- 1Vertical: −180 = 0 − ½(9.80)t², so t² = 36.73 and t = 6.06 s.
- 1Horizontal: x = 60.0 × 6.061 = 364 m ahead of the release point.
- 1Impact components: vx = 60.0 m/s and vy = −9.80 × 6.061 = −59.4 m/s.
- 1Speed = √(60.0² + 59.4²) = 84.4 m/s at tan−¹(59.4/60.0) = 44.7° below the horizontal.
Key terms
- Projectile
- A body moving freely under gravity alone after launch, so its acceleration is g downward and its path is a parabola.
- Range
- The horizontal distance between launch and landing points; on level ground it equals v0 squared times sin 2θ divided by g.
- Time of flight
- The interval between launch and landing, found from the vertical motion and then shared with the horizontal motion.
- Reference frame
- A viewpoint with its own origin and axes from which positions and velocities are measured, such as the ground or a moving car.
- Relative velocity
- A velocity measured from a moving frame, equal to the velocity of the object minus the velocity of that frame.
Projectile Motion and Relative Velocity FAQ
Why does the horizontal velocity of a projectile stay constant?
Once air resistance is ignored, the only force is gravity, and gravity acts straight down. Nothing pushes or pulls sideways, so the horizontal acceleration is zero and the horizontal component keeps its launch value until landing. Real objects slow a little because drag acts against the whole velocity.
Which root of the quadratic should I keep for the flight time?
Keep the positive time that lies after launch. A negative root describes when the object would have been at that height if the parabola were extended backwards, which never happened. If both roots are positive, the smaller one is the moment the object passes the height on the way up and the larger one on the way down.
Does the order matter in a relative velocity subtraction?
Yes. The velocity of A relative to B is the reverse of the velocity of B relative to A: same size, opposite direction. Write the quantity you want with subscripts first, then substitute, so the frame you are measuring from is always the velocity being subtracted. Mixing the order flips every direction in the answer.
Exam move
Work through projectiles in three passes. First, practise resolving launch velocities quickly, checking the sign of each component against a sketch. Second, take one launch and ask every standard question about it: time to the top, maximum height, flight time, range and impact velocity, using the same origin throughout.
Third, move the landing point above or below the launch point and solve the quadratic, deciding each time which root to keep. For relative motion, draw the vector triangle before calculating, label every arrow with its subscripts, and then repeat the answer in components to confirm it.
Finish with the first homework quiz, which covers Lectures 1 to 3, and enter answers to the number of significant figures common to the given values, because the quiz system checks that as well as the value.
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