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Motion in 2-D. Projectiles Circular Motion Updated 10.12.2014. Projectile Motion - what do you notice?. Two kinds of motion blended together. Vertical - acceleration Horizontal - constant velocity. Vertical d = v i t + ½ gt 2 vf = v i + gt vf = v i 2 +2gd. Horizontal V = d/t.

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## Motion in 2-D

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**Motion in 2-D**• Projectiles • Circular Motion • Updated 10.12.2014**Two kinds of motion blended together**• Vertical - acceleration • Horizontal - constant velocity**Vertical**d = vit + ½ gt 2 vf= vi + gt vf = vi2 +2gd Horizontal V = d/t The Return of the Big 5 (in part)**How long will it take a ball to fall when thrown at 20. m/s**horizontally from a height of 15 m? How far horizontally will it travel? • Draw the situation and label with given information • Decide upon formulae to use • Solve using formulae and showing UNITS!**Time to fall**d = vit + ½ gt 2 Solve for t Initial v in the vertical is 0 so t = sr (2d/g) T = sr (2(15m)/9.8m/s2) T = 1.7 s**How far horizontally will it travel?**d = vt d = 20m/s (1.7s) d = 34 m**A ball is launched at 4.5 m/s at 66° above the**horizontal.What are the maximum height and flight time of the ball? Establish a coordinate system with the initial position of the ball at the origin.**Show the positions of the ball at the beginning, at the**maximum height, and at the end of the flight.**Known Unknown**dyinit = 0 h max = dy = ? q = 66 o total flight time, t = ? vinit= 4.5 m/s a y = - g**Known Unknown**dyinit = 0 h max = dy = ? q = 66 o total flight time, t = ? vinit= 4.5 m/s a y = - g Find Vy using sin: Sin q = opp/hyp = Vy/4.5m/s Vy = 4.5m/s sin 66o Vy = 4.1m/s Vyf = Vi + gt Vyf = 0 (it stops) t = -Vi/-g = 4.1m/s/9.8m/s2 = t = 0.42 s Double this for up and down motion to get so 2 x 0.42 s total flight time = 0.84 s**Find the y-component of vi.**v yi = v i (sin q ) = 4.5 m/s (sin 66) = 4.1 m/s dy = ? dy = ½ g t2 = ½ (9.8 m/s2)(0.42s)2 dy = 0.86 m = 8.6 x 10-1 m**Ch 6: 2D Motion Assignment**• Projectiles: 1, 2, 3, 4, 51-53 Projectile Simulation http://www3.interscience.wiley.com:8100/legacy/college/halliday/0471320005/simulations6e/index.htm**Circular Motion**V An object moving in a circle at a constant speed is accelerated Centripetal acceleration depends upon the object’s speed and the radius of the circle Centripetal force causes centripetal acceleration. Fc ac Ch 6 Circular Motion Problems: 12-20, 49,61-63**Definitions**• Centripetal Force - a center seeking force that appears to pull an object toward the center of the circle along which it is moving • Centripetal Acceleration - acceleration toward the center ac = v2/r**Newton’s Law enters in…**• Since f = ma • The centripetal force is determined by substituting • ac = v2/r a = f/m • fc = mv2/r

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