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Projectile Motion Prior Knowledge Review and Reflect. Guided question: What concepts might we need to solve for the motion of a projectile launched at an angle, on the Earth? UM, UAM, Vectors, Resolving Vectors, Gravity, Free fall, Independence of Perpendicular Motions! Misconceptions: Diagnostics Two balls A and B fall freely, dropped simultaneously, with the height of ball A greater then the height of ball B. (a) Compare the velocity of the two balls when they reach the ground. (b) Compare the acceleration of the two balls. (c) Which ball will hit the ground first? Why? Misconceptions: Diagnostics A baseball is thrown at an angle from one person to another. (a) What is the velocity of the object at the peak of the flight? (b) What is the acceleration of the object at the peak of the flight? Misconceptions Ball A falls freely from a specific height and at the same time ball B is thrown horizontally from the same height. (a) Compare the velocity of the two balls when they reach the ground. (b) Compare the acceleration of the two balls. (c) Which ball will hit the ground first? Why? Misconceptions Applets or animated gifs are useful tools for visual learners. Here we see that heavier objects hit the ground at the same time as lighter ones. How might air resistance affect this? Common Misconceptions The parabolic motion of a projectile is different from one-dimensional motions. An object at rest drops to the ground before a similar object initially pushed horizontally falls to the ground. Heavier objects fall faster. At the top (peak) of its trajectory, the velocity of a projectile is always momentarily zero. The acceleration of the object is also zero, at the peak. Correcting Misconceptions Showing the parabolic nature of projectiles leads the student to believe they are quite different from linear falling or horizontal motion. An easy way to deal with this is to plot the motion of a freely falling object and a horizontally moving object on the same graph. When the intersection points at the same time, are connected, the result is a parabola. Correcting Misconceptions Have students look at the horizontal and vertical velocity vectors throughout parabolic flight. Demonstrate that the velocity is not zero as there must be some horizontal motion still (it doesn’t stop in midair). If you throw an object up in the air, it stops briefly (speed is zero) but acceleration is not zero (gravity doesn’t “shut off”). What is its role? To change the direction of the velocity vector. Is the velocity zero? Misconcpetions The green ball illustrates UM The red ball illustrates free fall and UAM The combination of these two motions at equal time intervals yields the parabolic path shown by the blue ball! Independence is clear. http://www.physicsclassroom.com/mmedia/vectors/bds.gif http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/mzs.gif Monkey and Banana Projectile Motion Treat 2 dimensional problems as a combination of two one-dimensional motions. In one dimension, the object undergoes UM. In the other dimension, the object undergoes UAM. We ignore air resistance. Problem Solving Subscripts are used to denote vertical and horizontal motions to alter our formulae as follows: sx = ux t UM sy = uy t + (½) at2 UAM vy = uy + at UAM Projectile Problems Each problem is solved as a system of equations in two independent motions. Both motions (UM and UAM) occur at the same time. (Thus Δt is exchangeable) Feel free to use the other UAM equations for problem solving. Become familiar with the quadratic equation. Problem #1 A purple dinosaur is dropped from a cliff 40. m high. How long does it take to strike the ground? As the motion is all vertical: sy = uyt + ½at2 - 40m = 0 + ½ (-9.81 m/s2) t2 t = 2.9 s Problem #2 A biology teacher is thrown at 10.0 m/s [E] from the top of a building that is 90.0 m tall. Calculate the horizontal displacement of the “test dummy” from the building upon initial contact with the ground. Step 1: Identify variables uy = 0, ux = +10.0 m/s, a = -9.81 m/s2, sy = - 90.0 m, t = ?, sx = ? Step 2:Choose formulae sx = uxΔt To solve for sx we need Δt. Use Vertical motion to find Δt sy = uyΔt + ½aΔt2 with uy = 0 - 90.0 m = ½ (-9.81 m/s2) t2 t = 4.2835 s Now sx = (10.0 m/s)(4.2835 s) = 42.8 m [E] Therefore, the annoying simpleton lands 42.8 m [E] of the base of the building. Links http://www.physicsclassroom.com/shwave /projectile http://phet.colorado.edu/sims/projectilemotion/projectile-motion_en.html Text work Page Page Page Page 59 63 67 69 C, D F #4 – 11 # 17 - 31