![Horizontal Direction
The initial velocity⃗
V0 can be decomposed into its x and y components using the launch
angle θ:
⃗
V0 =⃗
u0 +⃗
v0 = [V0 cos θ] x̂ + [V0 sin θ] ŷ
There is no horizontal acceleration (i.e. ax = 0), therefore the x-component of velocity
u(t) = u0 is constant. The kinematic equations reduce to a single equation:
x(t) = u0t = [V0 cos θ]t
where x is the horizontal position at time t
3](https://image.slidesharecdn.com/class04ap1physicslesson1-250225035716-fa4366c5/85/Class04_AP1_Physics_Lesson-1-doeiwjoejfo-pdf-4-320.jpg)
![Vertical Direction
There is a constant vertical acceleration due to gravity alone, i.e. ay = −g.1 The
important equation is this one:
y(t) = v0t −
1
2
gt2
= [V0 sin θ]t −
1
2
gt2
These two kinematic equations describing the y-component of velocity vy(t) may also
be useful:
vy = [V0 sin θ] − gt
v2
y =
V2
0 sin2
θ
− 2gy
1ay is negative due to the way we defined the coordinate system: the +y direction is up. Acceleration is
down, therefore negative.
4](https://image.slidesharecdn.com/class04ap1physicslesson1-250225035716-fa4366c5/85/Class04_AP1_Physics_Lesson-1-doeiwjoejfo-pdf-5-320.jpg)
![Example
Example: You are the pilot of a small plane and want to reach an airport, 600 km due
south, in 4.0 h. A wind is blowing at 50 km/h [S 35◦ E]2. With what heading and
airspeed should you fly to reach the airport on time?
2This is not a notation style that we use in AP Physics, but we should nevertheless be familiar with the
concepts presented in this example
18](https://image.slidesharecdn.com/class04ap1physicslesson1-250225035716-fa4366c5/85/Class04_AP1_Physics_Lesson-1-doeiwjoejfo-pdf-20-320.jpg)
Class04_AP1_Physics_Lesson (1)doeiwjoejfo.pdf
- 1. Class 4: Projectile Motion and Relative Motion AP Physics 1 Dr. Timothy Leung Winter 2025 Meritus Academy 1
- 3. Projectile Motion A projectile is an object that is launched with an initial velocity of⃗ V0 along a parabolic trajectory and accelerates only due to gravity. x y ⃗ V0 ⃗ v0 ⃗ u0 θ • x-axis: horizontal, pointing forward • y-axis: vertical, pointing up • Angle θ measured above the horizontal, i.e. • θ > 0 if above horizontal • θ < 0 if below horizontal • The origin is located at where the projectile is launched • Consistent with the right-handed Cartesian coordinate system 2
- 4. Horizontal Direction The initial velocity⃗ V0 can be decomposed into its x and y components using the launch angle θ: ⃗ V0 =⃗ u0 +⃗ v0 = [V0 cos θ] x̂ + [V0 sin θ] ŷ There is no horizontal acceleration (i.e. ax = 0), therefore the x-component of velocity u(t) = u0 is constant. The kinematic equations reduce to a single equation: x(t) = u0t = [V0 cos θ]t where x is the horizontal position at time t 3
- 5. Vertical Direction There is a constant vertical acceleration due to gravity alone, i.e. ay = −g.1 The important equation is this one: y(t) = v0t − 1 2 gt2 = [V0 sin θ]t − 1 2 gt2 These two kinematic equations describing the y-component of velocity vy(t) may also be useful: vy = [V0 sin θ] − gt v2 y = V2 0 sin2 θ − 2gy 1ay is negative due to the way we defined the coordinate system: the +y direction is up. Acceleration is down, therefore negative. 4
- 6. Solving Projectile Motion Problems Horizontal (x) and vertical (y) positions are independent of each other, but there are variables shared in both directions: • Time t • Launch angle θ (above the horizontal) • Initial speed V0 When solving any projectile motion problems • Two equations with two unknowns • If an object lands on an incline, there will be a third equation relating x and y 5
- 7. Projectile Motion Example Example: While hiking in the wilderness, you come to a cliff overlooking a river. A topographical map shows that the cliff is 291 m high and the river is 68.5 m wide at that point. You throw a rock directly forward from the top of the cliff, giving the rock a horizontal velocity of 12.8 m/s. (a) Did the rock make it across the river? (b) What is the speed of the rock when it reaches the bottom of the cliff? 6
- 8. Projectile Motion Example Example: A golfer hits the golf ball off the tee, giving it an initial velocity of 32.6 m/s at an angle of 65◦ with the horizontal. The green where the golf ball lands is 6.30 m higher than the tee, as shown below. (a) What is the time interval when the golf ball is in the air? (b) How far does the golf ball travel horizontally? 7
- 9. Projectile Motion Example Example: You are playing tennis with a friend on tennis courts surrounded by a 4.8 m fence. Your friend hits the ball over the fence, and you offer to retrieve it. You find the ball at a distance of 12.4 m on the other side of the fence. You throw the ball at an angle of 55.0◦ with the horizontal, giving it an initial velocity of 12.1 m/s. The ball is 1.05 m above the ground when you release it. Does the ball go over the fence, hit the fence, or hit the ground before reaching the fence? 8
- 10. Symmetric Trajectory A projectile’s trajectory is symmetric if the object lands at the same height as when it launched. The angle θ is measured above the horizontal. Time of flight: T = 2v0 sin θ g Horizontal range: R = v2 0 sin(2θ) g Maximum height: H = v2 0 sin2 θ 2g These equations are not given in the equation sheet for your AP exam, but they are relatively easy to derive. 9
- 11. Maximum Range R = v2 0 sin(2θ) g • Maximum range occurs at θ = 45◦ • For a given initial speed v0 and range R, launch angle θ is given by: θ1 = 1 2 sin−1 Rg v2 0 But there is another angle that gives the same range! θ2 = 90◦ − θ1 10
- 12. Relative Motion
- 13. Relative Motion All motion quantities must be measured relative to a frame of reference • Frame of reference: a coordinate system from which physical measurements are made • There is no absolute motion/rest: all motions are relative • Principle of relativity: All laws of physics are equal in all inertial (non-accelerating) frames 11
- 14. Relative Motion x′ C y′ x B y A The position and motion of A can be calculated from two frames of reference (coordinate systems) • B with axes x, y • C with axes x′, y′ The two reference frames may (or may not) be moving relative to each other. The motion of the two reference frames affect how motion of A is calculated. 12
- 15. Relative Motion x′ C y′ x B y A ⃗ rAC ⃗ rAB The position of A(t) can be described by • ⃗ rAB(t) (relative to B) • ⃗ rAC(t) (relative to C) Without needing mathematically rigorous vector notation, it is obvious that⃗ rAB(t) and⃗ rAC(t) are different vectors 13
- 16. Relative Motion x′ C y′ x B y ⃗ rBC A ⃗ rAC ⃗ rAB The position vector of the origins of the two reference frames is given by⃗ rBC(t) • The vector pointing from the origin of frame C to the origin of frame B • If the two frames are moving relative to each other, then⃗ rBC is also a function of time Without using vector notations, the relationship between the vectors is obvious: ⃗ rAC(t) =⃗ rAB(t) +⃗ rBC(t) 14
- 17. Relative Motion x′ C y′ x B y A ⃗ rAC ⃗ rAB ⃗ rBC Starting from the definition of relative position: ⃗ rAC =⃗ rAB +⃗ rBC Dividing by time, we get a similar equation for relative velocity: ⃗ vAC =⃗ vAB +⃗ vBC and relative acceleration: ⃗ aAC =⃗ aAB +⃗ aBC 15
- 18. Relative Velocity In classical mechanics, the equation for relative velocities follows the Galilean velocity addition rule, which applies to speeds much less than the speed of light: ⃗ vAC =⃗ vAB +⃗ vBC The velocity of A relative to reference frame C is the velocity of A relative to reference frame B, plus the velocity of B relative to C. If we add another reference frame D, the equation becomes: ⃗ vAD =⃗ vAB +⃗ vBC +⃗ vCD 16
- 19. Relative Velocity Example If an airplane (P) flies in windy air (A) we must consider the velocity of the airplane relative to air, i.e.⃗ vPA and the velocity of the air relative to the ground G, i.e.⃗ vAG. The velocity of the airplane relative to the ground is therefore ⃗ vPG =⃗ vPA +⃗ vAG Simple example: If an airplane is flying at a constant velocity of 253 km/h south relative to the air and the air velocity is 24 km/h east, what is the velocity of the airplane relative to Earth? 17
- 20. Example Example: You are the pilot of a small plane and want to reach an airport, 600 km due south, in 4.0 h. A wind is blowing at 50 km/h [S 35◦ E]2. With what heading and airspeed should you fly to reach the airport on time? 2This is not a notation style that we use in AP Physics, but we should nevertheless be familiar with the concepts presented in this example 18