pg 40 - 40 Chapter 3 Motion in Two and Three Dimensions...

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Unformatted text preview: 40 Chapter 3 Motion in Two and Three Dimensions Example 3.4 asked for the horizontal distance the car traveled. For that we needed the timeiwhich we weren’t given. This is a common situation in all but the simplest physics problems. You need to work through several steps to get the answer!in this case solving first for the unknown time and then for the distance. In essence, we solved two problems in Example 3.4: the first involving vertical motion and the second hori- l l l zontal motion. l i i; l - .1bq‘h-aw Instead of calculating t numerically in Example 3.4, we could have kept I: V 2y0/g in symbolic form until the end. That would avoid roundoff error and having to keep track of numerical digits and units. And you can often gain more physical insight from an answer that’s expressed symbolically before you put in the numbers. Projectile Trajectories g We’re often interested in the path, or trajectory, of a projectile without the details of where it is at each instant of time. We can specify the trajectory by giving the height y as a function of the horizontal position x. Consider a projectile launched from the origin at some angle 00 to the horizontal, with initial speed v0. As Fig. 3.16 suggests, the compo— nents of the initial velocity are raw: v0 cos 00 and vyo =v0 sin 60. Then Equations 3.12 and 3.13 become ' x = 12000560: and y = v0 sineor — £th Solving the x equation for the time I gives ElfillBflJfi Parabolic trajectory of a projectile. t = 1: v0 cos 90 Using this result in the y equation, we have 1‘ —vsin9 ———x el(’x 2 i i y 0 0 vocosfin 2g 11000560 1 f t i i i E 01' 8' . . , vi = x tant9 — ’72 r0 ectile tra'ector 3.14 y o 2%; 005260 (10 J J y) t ) Equation 3.14 gives a mathematical description of the projectile’s trajectory. Since )1 is a quadratic function of x, the trajectory is a parabola. However, there’s no new physics here, since Equation 3.14 follows directly from the constant-acceleration Equations 3.12 and 3.13. ' finding the Trajectorymut of the Hole A construction worker stands in a 2.6-rn-deep hole, 3.1 m from the ' _ ' - _ ' We want gothatfirehamer will ' . ' potter-m 3.1 m, edge of the hole, He tosses a hammer to’a'eonipani'on outside the hole. Ins-1 C153? If the hammer leaves his hand 1.0m above the-bottom of the hole at an- angle of 3-5“, what’s. the minimum speed it needs tie-clear the edge of the hole? How'farfrornthe edge of the hole does it; land? ; 5 integer: we’re concerned about where an object is but not when, so . we. mterpret this as a problem about the trajectory—specifically. the " ”minimum-speed trajectory that just grazes the edge of the hole. I. ' i mustar- We draw the Situation 'in Fig. 3.17} Equation 3.14 determines ; i :the trajectory. so our plan is to find the speed that makes the trajectory .thl‘O‘II,t}‘ifll4y§. . t ., pass just over the edge of the hole at x:3.i m, y= 1.6 In, Where Fig 3.1.7 shears- that We’ve chosen a coordinate system with its origin strum Our sketchfor Example 3.5. _ at'the worker‘s-hand. . - . . cont’dL ...
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