Topic 9: Motion in fields 9.1 Projectile motion 9.1.1 State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. 9.1.2 Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance. 9.1.3 Describe qualitatively the effect of air resistance on the trajectory of a projectile. 9.1.4 Solve problems on projectile motion. © 2006 By Timothy K. Lund

Topic 9: Motion in fields 9.1 Projectile motion State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. A projectile is an object that has been given an initial velocity by some sort of short-lived force, and then moves through the air under the influence of gravity. Baseballs, stones, or bullets are all examples of projectiles. You know that all objects moving through air feel an air resistance (recall sticking your hand out of the window of a moving car). © 2006 By Timothy K. Lund FYI We will ignore air resistance in the discussion that follows…

Topic 9: Motion in fields 9.1 Projectile motion State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. Regardless of the air resistance, the vertical and the horizontal components of velocity of an object in projectile motion are independent. © 2006 By Timothy K. Lund Slowing down in +y dir. Speeding up in -y dir. ay = -g ay = -g Constant speed in +x dir. ax = 0

Topic 9: Motion in fields 9.1 Projectile motion Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance. The trajectory of a projectile in the absence of air is parabolic. Know this! © 2006 By Timothy K. Lund

Topic 9: Motion in fields 9.1 Projectile motion Describe qualitatively the effect of air resistance on the trajectory of a projectile. If there is air resistance, it is proportional to the square of the velocity. Thus, when the ball moves fast its deceleration is greater than when it moves slow. SKETCH POINTS Peak to left of original one. © 2006 By Timothy K. Lund Pre-peak distance more than post-peak.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. Recall the kinematic equations from Topic 2: Since we worked only in 1D at the time, we didn’t have to distinguish between x and y in these equations. Now we appropriately modify the above to meet our new requirements of simultaneous equations: kinematic equations s = ut + (1/2)at2 v = u + at a is constant Displacement Velocity © 2006 By Timothy K. Lund kinematic equations ∆x = uxt + (1/2)axt2 vx = ux + axt ax and ay are constant ∆y = uyt + (1/2)ayt2 vy = uy + ayt

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. kinematic equations ∆x = uxt + (1/2)axt2 vx = ux + axt ax and ay are constant ∆y = uyt + (1/2)ayt2 vy = uy + ayt PRACTICE: Show that the reduced equations for projectile motion are SOLUTION: ax = 0 in the absence of air resistance. ay = -10 in the absence of air resistance. reduced equations of projectile motion ∆x = uxt vx = ux ∆y = uyt - 5t2 vy = uy - 10t © 2006 By Timothy K. Lund

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. reduced equations of projectile motion ∆x = uxt vx = ux ∆y = uyt - 5t2 vy = uy - 10t EXAMPLE: Use the reduced equations above to prove that projectile motion (in the absence of air resistance) is parabolic. SOLUTION: Just solve for t in the first equation and substitute it into the second equation. ∆x = uxt becomes t = ∆x/ux so that t2 = ∆x2/ux2. Then ∆y = uyt - 5t2, or ∆y = (uy/ux)∆x – (5/ux2)∆x2. © 2006 By Timothy K. Lund FYI The equation of a parabola is y = Ax + Bx2. In this case, A = uy/ux and B = -5/ux2.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. reduced equations of projectile motion ∆x = uxt vx = ux ∆y = uyt - 5t2 vy = uy - 10t PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (a) What are ux and uy? SOLUTION: Make a velocity triangle. u = 56 m s-1 © 2006 By Timothy K. Lund uy = u sin   = 15º uy = 56 sin 15º ux = u cos  uy = 15 m s-1. ux = 56 cos 15º ux = 54 m s-1

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. reduced equations of projectile motion ∆x = uxt vx = ux ∆y = uyt - 5t2 vy = uy - 10t PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (b) What are the tailored equations of motion? (c) When will the ball reach its maximum height? SOLUTION: (b) Just substitute ux = 54 and uy = 15: (c) At the maximum height, vy = 0. Why? Thus vy = 15 - 10t becomes 0 = 15 - 10t so that 10t = 15 t = 1.5 s. © 2006 By Timothy K. Lund tailored equations for this particular projectile ∆x = 54t vx = 54 ∆y = 15t - 5t2 vy = 15 - 10t

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. reduced equations of projectile motion ∆x = uxt vx = ux ∆y = uyt - 5t2 vy = uy - 10t PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (d) How far from the muzzle will the ball be when it reaches the height of the muzzle at the end of its trajectory? SOLUTION: From symmetry tup = tdown = 1.5 s so t = 3.0 s. Thus ∆x = 54t ∆x = 54(3.0) ∆x = 160 m. © 2006 By Timothy K. Lund

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. reduced equations of projectile motion ∆x = uxt vx = ux ∆y = uyt - 5t2 vy = uy - 10t PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (e) Sketch the following graphs: a vs. t, vx vs. t, and vy vs. t: SOLUTION: The only acceleration is g in the –y-direction. vx = 54, a constant. Thus it does not change over time. vy = 15 - 10t Thus it is linear with a negative gradient and it crosses the time axis at 1.5 s. © 2006 By Timothy K. Lund t ay -10 t vx 54 t vy 15 1.5

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. © 2006 By Timothy K. Lund The acceleration is ALWAYS g for projectile motion-since it is caused by Earth and its field. At the maximum height the projectile switches from upward to downward motion. vy = 0 at switch.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. The flight time is limited by the y motion. The maximum height is limited by the y motion. © 2006 By Timothy K. Lund

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. © 2006 By Timothy K. Lund ax = 0. ay = -10 ms-2.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. ∆y = uyt - 5t2 -33 = 0t - 5t2 -33 = -5t2 (33/5) = t2 © 2006 By Timothy K. Lund Fall time limited by y-equations: t = 2.6 s.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. ∆x = uxt ∆x = 18(2.6) © 2006 By Timothy K. Lund Use x-equations and t = 2.6 s: ∆x = 15 m.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. 18  vx = ux vx = 18. 26 vy = uy – 10t vy = 0 – 10t vy = –10(2.6) = -26. © 2006 By Timothy K. Lund tan  = 26/18  = tan-1(26/18) = 55º.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. © 2006 By Timothy K. Lund The horizontal component of velocity is vx = ux which is CONSTANT. The vertical component of velocity is vy = uy – 10t which is INCREASING (negatively).

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion.  ∆EK + ∆EP = 0 ∆EK = -∆EP ∆EK = -mg∆y EK0 = (1/2)mu2 © 2006 By Timothy K. Lund ∆EK = -(0.44)(9.8)(-32) = +138 J = EK – EK0 EK = +138 + (1/2)(0.44)(222) = 240 J.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. If 34% of the energy is consumed, 76% remains. 0.76(240) = 180 J (1/2)(0.44)v2 = 180 J © 2006 By Timothy K. Lund v = 29 ms-1.

Topic 9: Motion in fields 9.1 Projectile motion (1/2)mvf2 - (1/2)mv2 = -∆EP Topic 9: Motion in fields 9.1 Projectile motion mvf2 = mv2 + -2mg(0-H) vf2 = v2 + 2gH Solve problems on projectile motion. © 2006 By Timothy K. Lund Use ∆EK + ∆EP = 0.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. uy = u sin  uy = 28 sin 30º ux = u cos  ux = 28 cos 30º © 2006 By Timothy K. Lund ux = 24 m s-1. uy = 14 m s-1.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. ∆x = uxt 16 = 24t t = 16/24 = 0.67 The time to the wall is found from ∆x… © 2006 By Timothy K. Lund ∆y = uyt – 5t2 ∆y = 14t – 5t2 ∆y = 14(0.67) – 5(0.67)2 = 7.1 m.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. 0.5s © 2006 By Timothy K. Lund 0.0s 4 m ux = ∆x/∆t = (4-0)/(0.5-0.0) = 8 ms-1.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. 0.5s © 2006 By Timothy K. Lund 11 m 0.0s 4 m uy = ∆y/∆t = (11-0)/(0.5-0.0) = 22 ms-1.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. 2.5s 3.0s 2.0s 30 m 1.5s 1.0s D 0.5s © 2006 By Timothy K. Lund 11 m  = tan-1(30/24) = 51º 0.0s  4 m 24 m D2 = 242 + 302 so that D = 38 m ,@  = 51º.

Topic 9: Motion in fields 9.1 Projectile motion Solve problems on projectile motion. New peak below and left. Pre-peak greater than post-peak. © 2006 By Timothy K. Lund