NJC Projectile motion notes

National Junior College Science Department | Physics 1 5. PROJECTILE MOTION Content Page 5. Projectile Motion .......................................................................................................... 1 5.1 Free fall ..................................................................................................................... 2 5.1.1 What is weight? .................................................................................................. 2 5.1.2 Projectile Motion ................................................................................................. 3 5.2 Gravitational potential energy in a uniform field ........................................................ 9 5.3 Effects of air resistance ........................................................................................... 10 Learning Objectives 5.1 Free fall (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. 5.2 Gravitational potential energy in a uniform field (c) Derive, from the definition of work done by a force, the equation ߂ܧ௣=߂݃݉ℎ for gravitational potential energy changes in a uniform gravitational field (e.g. near the Earth’s surface). (d) Recall and use the equation ߂ܧ௣=߂݃݉ℎ to solve problems. 5.3 Effects of air resistance (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.

National Junior College Science Department | Physics 2 5.1 FREE FALL 5.1.1 What is weight? (a) Describe and use the concept of weight as the force experienced by a mass in a gravitational field. Weight is the force experienced by an object due to the gravitational attraction exerted on it by a massive body, such as the Earth. It can be described using the following formula: W = mg Where:  W is the weight of the object (in newtons, N),  m is the mass of the object (in kilograms, kg),  g is the gravitational field strength ( in N kg1 ), which is approximately 9.81  N kg1, near the Earth's surface. Question : Show gravitational field strength is numerically equal to acceleration of free fall. For an object to be in free fall, the only force acting on it is the gravitational force acting on the body or its weight. Resultant Force = Weight ma = mg a = g Where:  a is the acceleration of free fall. Therefore, gravitational field strength is numerically equal to acceleration of free fall. The concept of weight arises because gravity pulls objects toward the center of a massive body, like Earth. The force of gravity, on an object, depends on two things: 1. Mass of the object (m): The more massive an object, the greater the gravitational pull it will experience. 2. Gravitational field strength (g): The gravitational field strength at a point in a gravitational field is defined as the gravitational force exerted per unit mass acting on a small mass placed at that point. The weight of an object is a force, so it is measured in newtons (N), which are a unit of force in the International System of Units (SI). An object is said to be in free fall when the only force acting on it is the gravitational force (weight)

National Junior College Science Department | Physics 3 5.1.2 Projectile Motion Self-study resources A SLS lesson on Projectile Motion https://for.edu.sg/projectile- motion (b) Describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction. In this section we will study how an object moves in a plane under the action of a constant force. An example is the motion of a ball thrown at an angle to the vertical, as shown in Fig. 5.1. The constant force acting on it is its weight (or gravitational force). The path of the ball is a parabola. This type of motion is also known as projectile motion. Fig. 5.1 yor vertical direction x or horizontal direction

National Junior College Science Department | Physics 4 Analysis of a projectile motion Galileo first gave an accurate analysis of this motion by splitting the motion into its vertical and horizontal components, and consider these separately. The photograph in Fig. 5.2 was taken while a lamp emitted regular flashes of light. The ball on the left was dropped from rest and the other, a projectile, was thrown sideways at the same time. The images of each ball were captured at equal time interval apart.  The horizontal distance travelled by the projectile is equal at every time interval. So, the horizontal motion is due to uniform velocity.  Their vertical accelerations (due to gravity) are equal, showing that a projectile falls like a body which is dropped from rest. Its horizontal velocity does not affect its vertical motion. Therefore, the horizontal and vertical motions of a body are independent and can be treated separately. Fig. 5.2

National Junior College Science Department | Physics 5 Fig. 5.3 shows the trajectory (or path) of a particle (in black) projected with an initial velocity u at an angle θ to the horizontal from the origin. Air resistance is negligible. Fig. 5.3  The initial velocity can be resolved into its horizontal component ux and vertical component uy.  In the horizontal motion, the horizontal velocity vx = ux is constant throughout the motion.  In the vertical motion, it is equivalent to a free fall motion. Therefore, the vertical velocity vy changes its the magnitude and/or direction with constant acceleration g. At the highest point of the trajectory, vy = 0 (but the horizontal speed ≠ 0).

National Junior College Science Department | Physics 6 Steps to Solving Projectile Motion Problems Step 1 Resolve the initial velocity (of projection) into its horizontal and vertical components. Step 2 Determine the initial and final positions of the object for your calculation in step 3. Step 3 Select the equations of motion to determine the unknown value: Horizontal motion: sx = uxt Note: If two out of the three quantities are unknown, consider vertical motion. Vertical motion: vy = uy + ayt sy = uyt + 1 2ayt2 vy2 = uy2 + 2aysy Note: 1. Remember to assign either  or  positive when using these equations. 2. When the motion is in a uniform gravitational field, ay = g = 9.81 m s−2. Example 5.1 A ball is kicked from the top of a 60 m building. The initial velocity of the ball is 5.0 m s−1 at an angle to the horizontal of 60°. Air resistance is negligible. (a) The ball will reach a maximum height h before falling to the ground. Calculate (i) the time taken for the ball to reach h, and (ii) the height h. (b) When the ball to hit the ground, determine (i) the time from leaving the building to hitting the ground, (ii) the horizontal displacement x, and (iii) the velocity v of the ball just before it hits the ground. h 5.0 m s–1 60o x 60 m

National Junior College Science Department | Physics 7 Solution Resolve initial velocity: horizontal velocity ux = 5.0 cos 60° = 2.5 m s−1 vertical velocity uy = 5.0 sin 60° = 4.3 m s−1 (a) Consider the vertical motion of the ball from the point it is kicked to the point it reaches h. At the top of the trajectory, vertical velocity vy becomes 0. Take vectors directed upwards () as positive, uy = +5.0 sin 60° (), s = +h (), ay = −9.81 (), vy = 0, t = ? (i) Use vy = uy + ayt to determine t 0 = 5.0 sin 60° + (-9.81)t=20.766 ݐ 0.4414 = 0.441 s (ii) Use vy2 = uy2 + 2aysy to determine h 02 = (5.0 sin 60°)2 + 2(-9.81)h ℎ=0.9557=0.956 m Note: You can also use sy = uyt + 1 2ayt2 and the value of t in (i) to solve for h (b) (i