y=y (3.37)
0 +
1
2
(v 0 y+vy)t
vy=v 0 y−gt (3.38)
(3.39)
y=y 0 +v 0 yt−^1
2
gt^2
v (3.40)
y(^2) =v
0 y
(^2) − 2g(y−y
0 ).
Step 3.Solve for the unknowns in the two separate motions—one horizontal and one vertical.Note that the only common variable between themotions is timet. The problem solving procedures here are the same as for one-dimensionalkinematicsand are illustrated in the solved examples
below.Step 4.Recombine the two motions to find the total displacementsand velocityv. Because thex- andy-motions are perpendicular, we determine
these vectors by using the techniques outlined in theVector Addition and Subtraction: Analytical Methodsand employingA= Ax^2 +Ay^2 and
θ= tan−1(Ay/Ax)in the following form, whereθis the direction of the displacementsandθvis the direction of the velocityv:
Total displacement and velocity
(3.41)s= x^2 +y^2
θ= tan−1(y/x) (3.42)
(3.43)
v= vx^2 +vy^2
θ (3.44)
v= tan
−1(v
y/vx).
102 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS
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