JPJC 2026 Projectile Lecture Notes Students

Kinematics Lecture Note 2026/JPJC/PHYSICS/9478 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS Projectile Motion Content  Free fall  Gravitational potential energy in a uniform field  Effects of air resistance Learning Outcomes Candidates should be able to: (a) describe and use the concept of weight as the force experienced by a mass in a gravitational field (b) describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction (c) derive, from the definition of work done by a force, the equation ∆Ep = mg∆h for gravitational potential energy changes in a uniform gravitational field (e.g. near the Earth’s surface) (d) recall and use the equation ∆Ep = mg∆h to solve problems (e) describe qualitatively, with reference to forces and energy, the motion of bodies falling in a uniform gravitational field with air resistance, including the phenomenon of terminal velocity. 1 Two Dimensional Motion (b) Describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction. 1.1 Comparison between 1-D motion and 2-D motion Below shows the snapshots of three different types of motion: Path (snapshots of an object against a grid as background, at regular time intervals) Description Uniform velocity in x- direction No velocity and acceleration in y- direction. Uniform velocity in x- direction. Uniform velocity in y- direction. Uniform velocity in x- direction Uniform acceleration in y- direction Variables vx = constant ax = 0 vx = constant, vy = constant ax = 0 ay = 0 vx = constant, vy ≠ constant ax = 0 ay ≠ 0

Kinematics Lecture Note 2026/JPJC/PHYSICS/9478 2 An example of the third type of motion shown above is projectile motion, where an object moving in air (with no air resistance) has acceleration only in the vertical direction. The key idea in solving projectile motion problems is that the horizontal and vertical components of the motion can be treated independently. Kinematic quantities like displacement, velocity and acceleration along one direction is independent of kinematic quantities in the perpendicular direction. 1.2 Analysing Projectile Motion (a) Horizontal projection A cannon is shot out horizontally at a velocity u. There is no horizontal acceleration. If there was no gravitational force, it would be moving to point C in a horizontal line, where the cannon’s position is equally spaced at equal time interval. (xs ut , constant u) If there was no horizontal velocity, it would be dropping vertically to point A. The position/velocity of the cannon can be calculated using kinematic equations. ( 21 ,2 y ys gt v gt  , initial yu = 0 ). The actual path of a horizontally projected cannon ball is shown in the curved path to point B. The exact horizontal and vertical position at any time t is obtained by combining the positions along C and along A. Similarly, the velocity at any point is obtained by finding the vector sum of the horizontal and vertical velocities. The total time taken is calculated from the vertical components. The horizontal distance is calculated from the horizontal components. (b) Angled projection Consider an object projected in air at an angle of  above the horizontal. The path traced out by the object is parabolic in shape, as shown on the right: The initial velocity u can be resolved into 2 perpendicular components. Horizontal component is ux = u cos  and Vertical component is uy = u sin . Assume that air resistance is ignored,  there is no horizontal acceleration and therefore horizontal velocity ux is constant;  vertical acceleration is due to gravity only ( ay = g or ay = g), therefore vertical motion is uniformly accelerated;  the equations of motion can be applied to solve for the quantities in the x (horizontal) and y (vertical) directions;  time t is the same for both x, y components.  u  u uy = u sin  ux = u cos 

Kinematics Lecture Note 2026/JPJC/PHYSICS/9478 3 Example of a second-by-second exposure of an object projected at an angle to the horizontal is shown below. (Assume g = 10 m s2.) Common calculations: 1. Time of flight t: use the vertical component. ( 21 ,set 02 y y ys u t gt s   ) to find t. Can use y y upv u gt  to find upt . Total t = 2 upt 2. Horizontal range: use horizontal component. ( x xs u t ) 3. Velocity: combine both vertical and horizontal velocities to find the magnitude and direction. 222 yx vvv  The object’s velocity is v, at an angle of  below the horizontal. The direction of the velocity vector v is the tangent to the path at the point. 4. Maximum height: use the vertical component when final velocity = 0. The path is symmetrical about the highest point when there is no air resistance. The same set of kinematic equations in 1 dimension are still applicable. It becomes simpler for horizontal component. missing variable general equations of motion equations for projectile motion: horizontal component  xa 0 vertical component  ya 0 s atuv  xx uv  y y yv u a t  v 2 2 1 atuts  tus xx  2 2 1 tatus yyy  a tvus       2 tus xx  tvus yy y        2 t asuv 222  xx uv  yyyy sauv 2 22  10 m s2 vx  vy v v tan y x v v 

Kinematics Lecture Note 2026/JPJC/PHYSICS/9478 4 Problem solving strategy for Projectile Motion 1. Sketch the path of the projectile, including initial and final positions. 2. Resolve the initial velocity vector into x- and y-components. 3. Treat the horizontal motion and the vertical motion independently. 4. Time is the only quantity that is common for both x- and y- directions. 5. Follow the techniques for solving problems with constant velocity to analyse the horizontal motion of the projectile. 6. Follow the techniques for solving problems with constant acceleration to analyse the vertical motion of the projectile. (Note: vertical acceleration is 9.81 m s2 downwards) Example 1 A ball is projected from horizontal ground with an initial velocity of 15 1m s at an angle of 60 to the horizontal, as shown in the figure on the right. (a) Calculate, for this ball, the initial values of (i) the vertical component of the velocity, (ii) the horizontal component of the velocity. (b) Assuming that air resistance can be neglected, use your answers in (a) to determine (i) the maximum height H to which the ball rises, (ii) the time of flight, i.e. the time interval between the instant when the ball is projected and the instant when it returns to the ground level, (iii) the range, i.e. the displacement along the same horizontal level (distance between the point from which the ball is projected and the point where it strikes the ground). Take upward and rightward as positive (a) (i) uy = 15 sin 60 = 13 m s1 (ii) ux = 15 cos 60 = 7.5 m s1 (b) (i) vy2 = uy2 + 2aysy (ii) s = ut + 1 2 at2 (iii) R = uxt 0 = 132 + 2(9.81)H 0 = 13t + 1 2 (9.81)t2 = (7.5)(2.7) H = 8.6 m 4.905t – 13 = 0 = 20 m t = 2.7 s u = 15 m s1 Horizontal ground 60

Kinematics Lecture Note 2026/JPJC/PHYSICS/9478 5 Example 2 A tennis ball is thrown from a height of 1.0 m above the ground with a speed of 20 ms1 at an angle of 30 above the horizontal. On its downward motion, it strikes the top of a fence which is 4.0 m above the ground, as shown. Calculate (a) the greatest height above the ground reached by the ball, (b) the time taken by the ball to reach this height, (c) the time taken by the ball to reach the point of impact with the fence, (d) the ball’s velocity just before it hits the fence. Take upward and right