4 Tow-Dimensional Kinematics 4-1 Motion in Two Dimensions Recall the equation for motion in 1- dimension vf=vi+at xf=xi+1/2(vi+vf)t Δx=xf - xi Δx=vit+1/2at2 vav=(vi+vf)/2 vf2=vi2+2aΔx

In 2 dimension apply the equation of motion to x & y directions

vfx=vix+ax t vfy=viy+ay t xf=xix+1/2(vix+vfx)t yf=xiy+1/2(viy+vfy)t Equation of motion along x; Equation of motion along y vfx=vix+ax t vfy=viy+ay t xf=xix+1/2(vix+vfx)t yf=xiy+1/2(viy+vfy)t Δx=vixt+1/2ax t2 Δy=viyt+1/2ay t2 vfx2=vix2+2ax Δx vfy2=viy2+2ay Δy same time t !

Constant Velocity ax=0 ay=0 vfx=vix=vx vfy=viy=vy, Δx=vxt Δy=vyt Ex1. Ball velocity is 2 m/s, in a direction 300 with horizon, the ball travel 3 m along the x, find the displacement in y.

Constant Acceleration Ex. 4-2 Hummer Acceleration A hummingbird is flying in such a way that it is initially moving vertically with a speed of 4.6 m/s and accelerating horizontally at 11 m/s2 . Assuming the bird’s acceleration remains constant for the time interval of interest, find the horizontally and vertical distance through which it moves in 0.55 s.

4-2 Projectile Motion: Basic Equations Projectile motion, assumptions: Air resistance ignored The acceleration due to gravity is constant, downward, and has a magnitude equal to g=9.81 m/s2 The earth’s rotation is ignored. Projectile motion x-direction y-direction ax=0 ay=-g=-9.8 m/s2 xf=xi+vixt vyf=viy-gt vyf2=viy2-2gΔy

4-3 Zero Launch Angle constant velocity along x direction Sketch Choose coordinate system Known & unknown quantities in x & y direction Apply equations

The ball from a height 100 m with the 10 m/s horizontal velocity throws out, find how long the ball touch the ground and the horizontal displacement. y x Vix=10m/s xi xf h=100 m

Jumping a crevasse From 2. 75 m high, and jump 4 Jumping a crevasse From 2.75 m high, and jump 4.10 m the width of crevasse, find the minimal speed to land the other side. Find the land speed. vo 4.10 m 2.75 m

Parabolic Path x=vixt yf =yi+1/2ayt2=h+1/2(-g)t2= h-1/2gt2=h-(g/2vix2)x2 t=x/vix y=a+bx2