Chapter 4 Kinematics in 2D Motion in a plane vertical or horizontal But the motion in the x and y directions are independent except that they are coupled by the time Therefore we can break the problem into x and y parts We must use vectors displacement r x y velocity v vx vy acceleration a ax ay Usually r ay g y y r y x x Two Sets of Kinematic Equations 1 1 x f x i v xi v xf t f t i y f y i v yi v yf t f t i 2 2 1 2 x f x i v xi t f t i a x t f t i 2 1 2 v xf v xi a x t f t i 2 xf 2 xi v v 2a x x f x i y f y i v yi t f t i a y t f t i 2 v yf v yi a y t f t i 2 yf 2 yi v v 2a y y f y i We can solve problems using the same methods as for 1D but now we need to consider both x and y components simultaneously Example Motorcycle Jump Consider a motorcycle jumping between two buildings separated by a distance x The difference in heights of the buildings is y What initial velocity must the motorcycle have to just make it to the other building What is the time to cross to the other building What is the final velocity on impact y v i y x vxi vyi 0 ax 0 ay g x The x component vxf of the velocity remains constant throughout the flight time vxf vxi we neglect air resistance ax 0 therefore nothing to affect x motion Because of gravity once the motorcycle is in the air its speed in the y direction vyf increases from zero points down and therefore the height decreases The magnitude of vxi the resultant velocity vx1 vxi 1 also increases and the v1 angle of the resultant vy1 velocity vector with Vx2 vxi respect to the x axis 2 changes v2 v y2 x direction motion is the same as if motion occurred on a flat surface y direction motion is equivalent to dropping the motorcycle What about 2D motion in the horizontal plane No acceleration due to gravity Can rotate coordinate system to reduce problem to 1D if the motion is in a z straight line If motion has a curvature must be treated as 2D topic of 6 5 and ch 10 x v z x Return to Motorcycle Problem x direction 1 x vxi vxf 2 t vxi t t x vxi 2 Same information as first equation 3 vxf vxi since ax 0 4 Same information as third equation y direction 1 y vyf t 2 vyf 2 y t 2 y g t2 2 t 2 y g 3 vyf g t 4 v2yf 2g y three different ways to get vyf x 2 y t v xi g g v xi x 2 y What is the final velocity when motorcycle lands on the other roof We know from the 3rd x direction equation that vxf vxi Therefore we need only vyf From 4th ydirection equation 2 xf vf v v 2 yf 2 xi v 2 g y v xi 2 g y 1 2 v xi If y 0 or g 0 or vxi then vf vxi If y or g or vxi 0 then vf vyf Let s add some numbers x 50 0 ft y 20 0 ft g 32 2 ft s2 vxi x g 2 y 50 0 ft 32 2 ft s2 2 20 0 ft 44 9 ft s t x vxi 50 0 ft 44 9 ft s 1 11 s Or t 2 y g 2 20 0 ft 32 2 ft s2 1 11 s vyf g t 32 2 ft s2 1 11s 35 9 ft s Or vyf 2g y 2 32 2 ft s2 20 0 ft 35 9 ft s 2 xf vf v v 2 2 35 9 ft s 2 44 9 ft s yf y 57 5 ft s sin 1 vyf vf sin 1 35 9 57 5 38 6 vxf x vf 57 5 ft s 38 6 vyf vf Example Problem A soccer player kicks the soccer ball with an initial speed of 50 ft s at an angle of 37 with respect to the horizontal Find the maximum height of the ball s trajectory and the time it is at that point Given v0 50 ft s 0 37 vy1 0 Take t0 0 y0 0 Find t1 and y1 h Example Problem A skier leaves a ramp of a ski jump with a velocity of 10 0 m s 15 0 above the horizontal He lands on a slope of incline 50 0 Neglecting air resistance find a the distance from the ramp to where the jumper lands and b the velocity components just before landing
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