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‘haw 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 constantacceleration Equations 3.12
and 3.13. ' finding the Trajectorymut of the Hole A construction worker stands in a 2.6rndeep hole, 3.1 m from the ' _ '  _ ' We want gothatﬁrehamer will
' . ' potterm 3.1 m, edge of the hole, He tosses a hammer to’a'eonipani'on outside the hole. Ins1 C153?
If the hammer leaves his hand 1.0m above thebottom of the hole at an
angle of 35“, what’s. the minimum speed it needs tieclear 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 "
”minimumspeed 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‘shand. .  . . cont’dL ...
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 Spring '07
 KOPP
 Physics

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