General Physics 1: Projectile and Uniform Circular Motion PDF

General Physics 1 Motion in Space: Projectile Motion & Uniform Circular Motion MET: Translational Motion Projectile Motion HOW TO BE THE CHAMPION OF THE WORLD? Objectives Deduce the consequences of the independence of vertical and horizontal components of projectile motion. Calcu...

General Physics 1 Motion in Space: Projectile Motion & Uniform Circular Motion MET: Translational Motion Projectile Motion HOW TO BE THE CHAMPION OF THE WORLD? Objectives Deduce the consequences of the independence of vertical and horizontal components of projectile motion. Calculate range, time of flight, and maximum heights of projectiles. Differentiate uniform and non-uniform circular motion. Solve problems involving two-dimensional motion in contexts such as, but not limited to ledge jumping, movie stunts, basketball, safe locations during firework displays, and Ferris wheels. Infer quantities associated with circular motion such as tangential velocity, centripetal acceleration, tangential acceleration, radius of curvature. Assignment (Seatwork Notebook) In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 m from this point. If you toss the coin with a velocity of 6.4 m/s at an angle of 60° above the horizontal, the coin lands in the dish. You can ignore air resistance. (a) What is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish? y Trajectory 𝒗𝟎 𝒂 𝒂𝒙 = 𝟎, 𝒂𝒚 = −𝒈 x A projectile is any object with initial velocity and followed a path generated by the effect of gravitational acceleration and air resistance.  Assumption: 1. Represent each projectile/object as a point particle with an acceleration, due to gravity, that is constant for both magnitude and direction. 2. Ignore the effect of air resistance and the curvature of the earth. If you have two objects both at the same initial point. One ball is dropped and the other is given a horizontal launch. Neglecting air resistance, which of the two reaches the floor first?  t=0.0 s t=1.0 s t=2.0 s t=3.0 s t=4.0 s y v0 x y x y x y x y x y The horizontal and vertical motions are independent of each other Both motions are of equal time (t) x Cases of Projectile Motion 𝒂𝒙 = 𝟎, 𝒂𝒚 = −𝒈 Case 1: Thrown at an angle PROJECTILE MOTION: In projectile motion with no air resistance, 𝑎𝑥 = 0 and 𝑎𝑦 = −𝑔. The coordinates and velocity components are simple functions of time, and the shape of the path is always a parabola. We usually choose the origin to be at the initial position of the projectile. Horizontal Motion: 1 𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡 𝑥 = 𝑥0 + 𝑣0𝑥 𝑡 + 𝑎𝑥 𝑡 2 2 = 𝑣0𝑥 + (0)(0) 1 𝒗𝒙 = 𝒗𝟎𝒙 = 0 + 𝑣0𝑥 𝑡 + (0)(0)2 2 𝒙 = 𝒗𝟎𝒙 𝒕 Vertical Motion: 1 𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡 𝑦 = 𝑦0 + 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 2 𝑜𝑟 1 𝒗𝒚 = 𝒗𝟎𝒚 − 𝒈𝒕 = 0 + 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 2 𝟏 𝒚 = 𝒗𝟎𝒚 𝒕 + 𝒂𝒚 𝒕𝟐 𝟐 Maximum Height, h For maximum height, h, 𝟏 𝒉 = 𝒗𝟎𝒚 𝒕 + 𝒂𝒚 𝒕𝟐 (𝑖𝑓 𝑡𝑖𝑚𝑒 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛) 𝟐 1 𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡 ℎ = 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 2 2 0 = 𝑣0𝑦 + 𝑎𝑦 𝑡 −𝑣0𝑦 1 −𝑣0𝑦 = 𝑣𝑜𝑦 + 𝑎𝑦 𝑎𝑦 2 𝑎𝑦 −𝑣0𝑦 𝑡= (𝟏) −𝑣0𝑦 2 1 𝑣0𝑦 2 𝑎𝑦 = + 𝑎𝑦 2 𝑎𝑦 1 −(𝒗𝟎𝒚 )𝟐 𝑦 = 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 (𝟐) 𝒉= 𝑖𝑓 𝑣0𝑦 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛, 𝑎𝑛𝑑 𝒕 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 2 𝟐𝒂𝒚 Maximum Height, h −(𝒗𝟎𝒚 )𝟐 𝒉= 𝑖𝑓 𝑣0𝑦 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛, 𝑎𝑛𝑑 𝒕 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡 𝟐𝒂𝒚 0 = 𝑣0𝑦 + 𝑎𝑦 𝑡 𝑡= −𝑣0𝑦 (𝟏) Where, 𝑣𝑜𝑦 = sin 𝜃 𝑣0 , 𝑎𝑦 1 −(𝑣0 𝑠𝑖𝑛𝜃)2 𝑦 = 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 (𝟐) ℎ= 2 2𝑎𝑦 −(𝒗𝟎 𝟐 𝒔𝒊𝒏𝟐 𝜽) 𝒉= 𝑖𝑓 𝑣0𝑦 𝑎𝑛𝑑 𝒕 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝟐𝒂𝒚  𝑥 = 𝑣0𝑥 𝑡 𝑹 = 𝒗𝟎𝒙 𝒕𝟐 = 𝑣0𝑥 2𝑡1 𝑹 = 𝒗𝟎𝒙 𝟐𝒕𝟏 −𝑣0𝑦 We can also substitute the expression, 𝑡 = 𝑎𝑦 𝑅 = 𝑣0𝑥 2𝑡1 −𝒗𝟎𝒚 𝑹 = 𝒗𝟎𝒙 𝟐( ) 𝒂𝒚  −𝑣0𝑦 We can also substitute the expression, 𝑡 = 𝑎𝑦 Recall: Trigonometric Identity 𝑅 = 𝑣0𝑥 2𝑡1 2sinθcosθ=sin2θ −𝒗𝟎𝒚 𝑹 = 𝒗𝟎𝒙 𝟐( ) 𝒂𝒚 Let’s substitute the value for v0x =𝑣0 𝑐𝑜𝑠𝜃, and v0y==𝑣0 sin 𝜃 to the eq.above: −𝑣0𝑦 𝑅 = 𝑣0𝑥 2 𝑎𝑦 −(𝑣0 𝑠𝑖𝑛𝜃) = 2(𝑣0 𝑐𝑜𝑠𝜃) 𝑎𝑦 −(𝒗𝟎 𝟐 𝐬𝐢𝐧 𝟐𝜽) 𝑹= 𝒂𝒚 Example 1: Analyzing Projectile Motion Carlos hit the soccer ball to his teammates. It leaves at a speed 𝑣0 = 25.6 𝑚/𝑠 at an angle 𝜃0 = 67.30. Ignoring air resistance, find the following: (a)The coordinates of the position of the ball and its velocity (magnitude and direction) at t=1.50 s. (b) Find the time when the ball reaches the highest point of its flight, and its height h at this time. (c)Find the horizontal range R covered by the soccer ball. Given: 𝑣0 = 25.6 𝑚/𝑠 𝜃0 = 67.30 Find: (a) x, y, and 𝒗𝟏.𝟓𝟎 (b)𝑡1 , h (c) R  (a) First, calculate the values of v0x and voy Example 1: Analyzing Projectile Motion 𝑣0𝑦 = 𝑣0 𝑠𝑖𝑛𝜃 𝑚 = 25.6 𝑠𝑖𝑛67.30 𝑠 𝒗𝟎𝒚 = 𝟐𝟑. 𝟔𝟐 𝒎/𝒔 For x-coordinate: 𝑥 = 𝑣0𝑥 𝑡 𝑚 = 9.879 (1.50 𝑠) 𝑠 Given: 𝒙 = 𝟏𝟒. 𝟖 𝒎 𝑣0 = 25.6 𝑚/𝑠 For y-coordinate: 𝜃0 = 67.30 1 𝑦 = 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 2 𝑚 1 𝑚 = 23.62 1.50 𝑠 + (−9.8 2 )(1.50 𝑠)2 𝑠 2 𝑠 = 35.43 𝑚 − 11. 025𝑚 𝒚 = 𝟐𝟒. 𝟒𝟏 𝒎 Example 1: Analyzing Projectile Motion (a) Calculate the values of vx and vy 𝑣𝑥 = 𝑣0𝑥 𝒗𝒙 = 𝟗. 𝟕𝟖𝟗 𝒎/𝒔 𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡 𝑚 𝑚 = 23.62 + −9.8 2 1.50 𝑠 𝑠 𝑠 Given: 𝑣0 = 25.6 𝑚/𝑠 𝑚 𝑚 𝜃0 = 67.30 = 23.62 − 14.7 𝑠 𝑠 𝒎 𝒗𝒚 = 𝟖. 𝟗𝟏𝟕 𝒔 Example 1: Analyzing Projectile Motion (a) Then, let us use Pythagorean theorem to get the magnitude and calculate for direction. 𝑣= 𝑣𝑥 2 + 𝑣𝑦 2 𝑚 2 𝑚 2 = 9.879 + 8.916 Given: 𝑠 𝑠 𝑣0 = 25.6 𝑚/𝑠 𝜃0 = 67.30 𝒎 𝒗= 𝟏𝟑. 𝟑𝟏 𝒔  Example 1: Analyzing Projectile Motion (a) To calculate for direction, 𝑣𝑦 𝜃 = 𝑎𝑟𝑐 tan 𝑣𝑥 𝑚 8.916 = 𝑎𝑟𝑐 tan 𝑠 𝑚 Given: 9.879 𝑠 𝑣0 = 25.6 𝑚/𝑠 𝜃0 = 67.30 𝜽 = 𝟒𝟐. 𝟎𝟕𝟎 Thus, at t=1.50 s, the soccer ball is at (14.8 m, 24.41 m) with a velocity vector, 13.31 m/s at 42.070.  (b) Finding 𝑡1 , h Example 1: Analyzing Projectile Motion 𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡1 = 0 𝑚 −𝑣0𝑦 −(23.62 ) 𝑡1 = = 𝑠 = 𝟐. 𝟒𝟏𝟎 𝒔 𝑎𝑦 −9.8 𝑚/𝑠 2 Since, we already have the have for time to reach the maximum height, 1 ℎ = 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 Given: 2 𝑣0 = 25.6 𝑚/𝑠 𝑚 1 𝑚 𝜃0 = 67.30 = 23.62 2.410 𝑠 + (−9.8 2 )(2.410𝑠)2 𝑠 2 𝑠 = 56.92 𝑚 − 28.46 𝑚 𝒉 = 𝟐𝟖. 𝟒𝟔 𝒎 Thus, it takes 2.410 s to reach 28.46 m – the maximum height reached by the soccer ball. Example 1: Analyzing Projectile Motion (c) Finding R To calculate for range, R, let’s find first the value for t2. Since our soccer ball already hit the ground at t2, we have now y = 0. 1 𝑦 = 𝑣0𝑦 𝑡2 + 𝑎𝑦 𝑡2 2 = 0 2 1 𝑡2 𝑣0𝑦 + 𝑎𝑦 𝑡2 = 0 2 Given: This is a quadratic equation and it has two roots; 𝑣0 = 25.6 𝑚/𝑠 𝜃0 = 67.30 𝑡2 = 0; 𝑚 −2𝑣0𝑦 −(2) 23.62 𝑠 𝑡2 = = 𝑚 𝑎𝑦 −9.8 2 𝑠 𝒕𝟐 = 𝟒. 𝟖𝟐𝟎 𝒔  (c) Finding R Example 1: Analyzing Projectile Motion substitute the value for t2 in eq. for R, 𝑅 = 𝑣0𝑥 𝑡2 𝑚 = 9.879 4.820 𝑠 𝑠 Given: 𝑹 = 𝟒𝟕. 𝟏𝟖 𝒎 𝑣0 = 25.6 𝑚/𝑠 𝜃0 = 67.30 Thus, the soccer ball covered a range of 47.18 m after being kicked by Carl. Case 2: Thrown Horizontally Example 2: Analyzing Projectile Motion Mark and his friend went to Canibad to enjoy the summer and try cliff diving. In 53.0 m high cliff with respect to the surface of the sea water, Mark dives horizontally with a speed of 32.5 m/s. a) How long does he remained in the air before hitting the sea water? b) At what horizontal distance from the diving point does he strike the sea water? c) What is the vertical component of velocity as he strikes the surface of sea water? Given: 𝑣0 = 32.5 𝑚/𝑠 𝑦 = −53.0 𝑚 Find: (a) t (b) R (c) 𝑣𝑦  Example : Analyzing Projectile Motion (a) Let’s evaluate first the vertical motion in this problem. Take note that the initial vertical position of the projectile y0 =0 and the final position y= -53.0 m, since it hits the surface already. 1 𝑦 = 𝑦0 + 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 2 1 𝑦 − 𝑦0 = 0 𝑡 + 𝑎𝑦 𝑡 2 2 1 𝑦 = 𝑎𝑦 𝑡 2 2 Given: Find: (a) t 𝑖𝑠𝑜𝑙𝑎𝑡𝑒 𝒕 𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑒 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑣0 = 32.5 𝑚/𝑠 𝑦 = −53.0 𝑚 (b) R (c) 𝑣𝑦 1 2 𝑦 = 𝑎𝑦 𝑡 2 2 2 2𝑦 𝑎𝑦 𝑡 2 = 𝑎𝑦 𝑎𝑦 Example 2: Analyzing Projectile Motion (a) 2𝑦 𝑎𝑦 𝑡 2 = 𝑎𝑦 𝑎𝑦 2𝑦 𝑡= 𝑎𝑦 2 −53.0 𝑚 𝑡= 𝑚 = 𝟑. 𝟐𝟗 𝒔 −9.8 2 𝑠 Find: Thus, Mark stays in the air for 3.29 s before hitting the Given: (a) t surface of sea water. 𝑣0 = 32.5 𝑚/𝑠 (b) R 𝑦 = −53.0 𝑚 (c) 𝑣𝑦 Example 3: Analyzing Projectile Motion (b) Finding R The value of our initial horizontal position is 𝑥0 = 0 𝑚. Let, x = R 1 𝑅 = 𝑥0 + 𝑣𝑜𝑥 𝑡 + 𝑎𝑥 𝑡 2 2 𝑚 1 2 𝑅 = 0 𝑚 + 32.5 3.29 𝑠 + 0 3.29 𝑠 𝑠 2 𝑹 = 𝟏𝟎𝟕 𝒎 Find: Given: (a) t 𝑣0 = 32.5 𝑚/𝑠 Thus, Mark covered 107 m is the horizontal distance from the (b) R 𝑦 = −53.0 𝑚 diving point to where he strikes the sea water. (c) 𝑣𝑦 Example 3: Analyzing Projectile Motion (C) Find vy, 𝑣𝑦 = 𝑣𝑜𝑦 + 𝑎𝑦 𝑡 𝑚 𝑣𝑦 = 0 + −9.8 2 3.29 𝑠 𝑠 𝒗𝒚 = −𝟑𝟐. 𝟐 𝒎/𝒔 Thus, the vertical component of the velocity as Mark hits the Find: surface of sea water is equal to -32.2 m/s. Given: (a) t 𝑣0 = 32.5 𝑚/𝑠 (b) R 𝑦 = −53.0 𝑚 (c) 𝑣𝑦 Example 2: Analyzing Projectile Motion At the edge-top of a 78.0 m tall building, a janitor throws a stone at an angle 𝜃0 = 78. 00 and it reaches a maximum height of 10.0 m with respect to the rooftop. Find the following: (a) What would be the minimum speed of the stone to reach the maximum height? (b) How far from the building did the stone landed with respect to the ground? (c) How long does the stone in the air before reaching the ground? PROJECTILE MOTION: In projectile motion with no air resistance, 𝑎𝑥 = 0 and 𝑎𝑦 = −𝑔. The coordinates and velocity components are simple functions of time, and the shape of the path is always a parabola. We usually choose the origin to be at the initial position of the projectile. Uniform Circular Motion When an object is moving in a circular path with constant speed, that is called uniform circular motion Δ𝑣Ԧ Δ𝑠 𝑣1 = 𝑜𝑟 Δ𝑣 = ∆𝑠 𝑣1 𝑅 𝑅 The magnitude of average acceleration 𝑎Ԧ𝑎𝑣 during time interval ∆𝑡; |∆𝑣| 𝑣1 ∆𝑠 𝑎𝑎𝑣 = = ∆𝑡 𝑅 ∆𝑡 The magnitude of instantaneous acceleration as the particle move from P1 to P2 for the time interval, 𝑣1 ∆𝑠 𝑎 = lim ∆𝑡→0 𝑅 ∆𝑡  𝑣1 ∆𝑠 𝑎 = lim ∆𝑡→0 𝑅 ∆𝑡 Centripetal Acceleration/Radial Acceleration 𝒗𝟐 𝒂𝒓𝒂𝒅 = 𝑹 Where, arad = magnitude centripetal acceleration or radial acceleration 𝑣 2 = the speed of object 𝑅= radius of object’s circular path Centripetal Acceleration/Radial Acceleration 𝒗𝟐 𝒂𝒓𝒂𝒅 = 𝑹 𝟐𝝅𝑹 𝟏 𝒗= , 𝑻= 𝑻 𝒇 So, we can rewrite our formula for arad by substituting the value of v; 𝟒𝝅𝟐 𝑹 𝒂𝒓𝒂𝒅 = 𝑻𝟐 Uniform Circular Motion vs. Non-Uniform Circular Motion If the speed of an moving object in a circular motion varies, we call it nonuniform circular motion. It has parallel component 𝑎𝑡𝑎𝑛 , is still tangent to the path and parallel to the velocity 𝑣Ԧ of the object. It has perpendicular component or radial acceleration 𝑎𝑟𝑎𝑑 , that is directed toward the center. 𝒗2 𝒅|𝒗| 𝒂𝒓𝒂𝒅 = 𝒂𝒏𝒅 𝒂𝒕𝒂𝒏 = 𝑹 𝒅𝒕 (𝑛𝑜𝑛 − 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛) Example 4: Analyzing Uniform Circular Motion An 85-kg halfback makes a turn in a football field. The halfback sweeps out a curve path that is part of a circle with a radius of 25 m. It then takes a quarter of a turn around the circle in 3.5 s. What is the speed and acceleration of the halfback? Would there be change in the speed of the halfback after 3.5 s? Given: Solution: m = 85 kg Finding 𝒗, R = 25 m 𝑑 t = 3.5 s 𝑣= 𝑡 1 Find: 2𝜋𝑅 = 4 𝒗=? 𝑡 1 𝒂𝒓𝒂𝒅 = ? 4 [2 3.14 25 𝑚 ] = 3.5 𝑠 𝒗 = 𝟏𝟏. 𝟐 𝒎/𝒔 Example 4: Analyzing Uniform Circular Motion An 85-kg halfback makes a turn in a football field. The halfback sweeps out a curve path that is part of a circle with a radius of 25 m. It then takes a quarter of a turn around the circle in 3.5 s. What is the speed and acceleration of the halfback? Would there be change in the speed of the halfback after 3.5 s? Given: Solution: m = 85 kg Finding𝒂𝒓𝒂𝒅 , R = 25 m 𝑣2 t = 3.5 s 𝑎𝑟𝑎𝑑 = 𝑅 𝑚 (11. 2 𝑠 )2 Find: = 25 𝑚 𝒗=? 𝒎 𝒂𝒓𝒂𝒅 = ? 𝑎𝑟𝑎𝑑 = 𝟓. 𝟎𝟐 𝒔𝟐 Seatwork Seatwork Notebook Example 5: Analyzing Uniform Circular Motion A racecar is driving along the racetrack, with a tangential speed of 30 m/s. It takes 53 seconds to make one loop. (a)What is the radial acceleration of the racecar? (b)If the speed doubles and the radial acceleration is 5.00 m/𝑠 2 , how many seconds will the racecar takes to make one loop? Given: Find: v = 30.0 m/s 𝒂𝒓𝒂𝒅 = ? T = 53 s t=? Example 5: Analyzing Uniform Circular Motion Combine (1) and (2), to derive A racecar is driving along the racetrack, with a tangential speed of the formula for 𝑎𝑟𝑎𝑑 , 30 m/s. It takes 53 seconds to make one loop. (a) What is the radial acceleration of the racecar? 𝑣2 (b) If the speed doubles and the radial acceleration is 5.00 m/𝑠 2 , 𝑎𝑟𝑎𝑑 = how many seconds will the racecar takes to make one loop? 𝑅 𝑣2 Solution: = 𝑣𝑇 a) Find 𝑎𝑟𝑎𝑑 =? 2𝜋 𝑣 𝑣2 2𝜋𝑅 𝑎𝑟𝑎𝑑 = Recall: 𝑎𝑟𝑎𝑑 = (1) , 𝑣 = 𝑇 𝑅 𝑇 2𝜋 From the formula for v, let’s derive the 30.0 𝑚/𝑠 formula for R, = 53.00 𝑠 𝑣𝑇 = 2𝜋𝑅 2𝜋 𝑣𝑇 2𝜋𝑅 𝒎 = 𝒂𝒓𝒂𝒅 = 𝟑. 𝟓𝟓𝟕 𝟐 2𝜋 2𝜋 𝒔 𝑣𝑇 𝑅= (2) 2𝜋 Example 5: Analyzing Uniform Circular Motion 𝑇 A racecar is driving along the racetrack, with a tangential speed of 𝑎𝑟𝑎𝑑 =𝑣 2𝜋 30 m/s. It takes 53 seconds to make one loop. 𝑎𝑟𝑎𝑑 𝑇 (a) What is the radial acceleration of the racecar? =𝑣 (b) If the speed doubles and the radial acceleration is 6.50 m/𝑠 2 , 2𝜋 how many seconds will the racecar takes to make one loop? 𝑣2𝜋 = 𝑎𝑟𝑎𝑑 𝑇 𝑣2𝜋 Solution: 𝑇= 𝑎𝑟𝑎𝑑 b) Find 𝑇 =? 𝑚 (60.0 𝑠 )(2)(𝜋) From the working equation for 𝑎𝑟𝑎𝑑 , = 𝑚 let’s derive the formula for T, 6.50 2 𝑠 𝑣 𝑇 = 58.00 𝑠 𝑎𝑟𝑎𝑑 = 𝑇 2𝜋 𝑇 𝑎𝑟𝑎𝑑 =𝑣 2𝜋 UNIFORM CIRCULAR MOTION: When a particle moves in a circular path of radius R with constant speed v (uniform circular motion), it’s acceleration is directed toward the center of the circular perpendicular to 𝒗. The magnitude 𝒂𝒓𝒂𝒅 of the acceleration can be expressed in terms of v and R or in terms of R and the period T. If the speed is not constant in circular motion (nonuniform circular motion), there is still a radial component of 𝒂, but there are also a component of 𝒂 parallel (tangential) to the path. This tangential component is equal to the rate of change of speed. Reflection In the conquest of humankind to understand the universe and look for livable planet, what is the relevant connection and importance of the topics (Projectile and Uniform Circular Motion) ?