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9.1.1State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. 9.1.2Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance. 9.1.3Describe qualitatively the effect of air resistance on the trajectory of a projectile. 9.1.4Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion
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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). Topic 9: Motion in fields 9.1 Projectile motion FYI We will ignore air resistance in the discussion that follows…
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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. Topic 9: Motion in fields 9.1 Projectile motion Slowing down in +y dir. Speeding up in -y dir. Constant speed in +x dir. a x = 0 a y = -g
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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! Topic 9: Motion in fields 9.1 Projectile motion
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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. Topic 9: Motion in fields 9.1 Projectile motion SKETCH POINTS Peak to left of original one. Pre-peak distance more than post-peak.
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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: Topic 9: Motion in fields 9.1 Projectile motion kinematic equations s = ut + (1/2)at 2 v = u + at a is constant Displacement Velocity kinematic equations ∆x = u x t + (1/2)a x t 2 v x = u x + a x t a x and a y are constant ∆y = u y t + (1/2)a y t 2 v y = u y + a y t
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion kinematic equations ∆x = u x t + (1/2)a x t 2 v x = u x + a x t a x and a y are constant ∆y = u y t + (1/2)a y t 2 v y = u y + a y t PRACTICE: Show that the reduced equations for projectile motion are SOLUTION: a x = 0 in the absence of air resistance. a y = -10 in the absence of air resistance. 0 0 reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 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 = u x t becomes t = ∆x/u x so that t 2 = ∆x 2 /u x 2. Then ∆y = u y t - 5t 2, or ∆y = (u y /u x )∆x – (5/u x 2 )∆x 2. FYI The equation of a parabola is y = Ax + Bx 2. In this case, A = u y /u x and B = -5/u x 2. reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 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 u x and u y ? SOLUTION: Make a velocity triangle. u = 56 m s -1 u x = u cos u y = u sin = 15º u x = 56 cos 15º u x = 54 m s -1 u y = 56 sin 15º u y = 15 m s -1.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 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 u x = 54 and u y = 15: (c) At the maximum height, v y = 0. Why? Thus v y = 15 - 10t becomes 0 = 15 - 10t so that 10t = 15 t = 1.5 s. tailored equations for this particular projectile ∆x = 54t v x = 54 ∆y = 15t - 5t 2 v y = 15 - 10t
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 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 t up = t down = 1.5 s so t = 3.0 s. Thus ∆x = 54t ∆x = 54(3.0) ∆x = 160 m.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 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, v x vs. t, and v y vs. t: SOLUTION: The only acceleration is g in the – y-direction. v x = 54, a constant. Thus it does not change over time. v y = 15 - 10t Thus it is linear with a negative gradient and it crosses the time axis at 1.5 s. t ayay -10 t vxvx 54 t vyvy 15 1.5
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 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. v y = 0 at switch.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion The flight time is limited by the y motion. The maximum height is limited by the y motion.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion a x = 0. a y = -10 ms -2.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion Fall time limited by y-equations: ∆y = u y t - 5t 2 -33 = 0t - 5t 2 -33 = -5t 2 (33/5) = t 2 t = 2.6 s.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion Use x-equations and t = 2.6 s: ∆x = u x t ∆x = 18(2.6) ∆x = 15 m.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion v x = u x v y = u y – 10t v x = 18. v y = 0 – 10t v y = –10(2.6) = -26. 18 26 tan = 26/18 = tan -1 (26/18) = 55º.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion The horizontal component of velocity is v x = u x which is CONSTANT. The vertical component of velocity is v y = u y – 10t which is INCREASING (negatively).
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion ∆E K + ∆E P = 0 ∆E K = -∆E P ∆E K = -mg∆y ∆E K = -(0.44)(9.8)(-32) = + 138 J = E K – E K0 E K = + 138 + (1/2)(0.44)(22 2 ) = 240 J. E K0 = (1/2)mu 2
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion If 34% of the energy is consumed, 76% remains. 0.76(240) = 180 J (1/2)(0.44)v 2 = 180 J v = 29 ms -1.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion Use ∆E K + ∆E P = 0. (1/2)mv f 2 - (1/2)mv 2 = -∆E P mv f 2 = mv 2 + -2mg(0-H) v f 2 = v 2 + 2gH
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion u x = u cos u x = 28 cos 30º u x = 24 m s -1. u y = u sin u y = 28 sin 30º u y = 14 m s -1.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion ∆x = u x t 16 = 24t t = 16/24 = 0.67 ∆y = u y t – 5t 2 ∆y = 14t – 5t 2 ∆y = 14(0.67) – 5(0.67) 2 = 7.1 m. The time to the wall is found from ∆x…
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 0.0s 0.5s 4 m u x = ∆x/∆t = (4-0)/(0.5-0.0) = 8 ms -1.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 0.0s 0.5s 4 m 11 m u y = ∆y/∆t = (11-0)/(0.5-0.0) = 22 ms -1.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 0.0s 0.5s 4 m 11 m 1.0s 1.5s 2.0s 2.5s 3.0s 24 m 30 m D 2 = 24 2 + 30 2 so that D = 38 m D = tan -1 (30/24) = 51º,@ = 51º.
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Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion New peak below and left. Pre-peak greater than post- peak.
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