PHYS 2010 Nathalie Hoffmann University of Utah Tuesday 5/26 PHYS 2010 Nathalie Hoffmann University of Utah

2D Kinematics (Yay!) Acceleration Same kinematics equations Component in direction of motion -> speeding up or slowing down Component perpendicular to direction of motion -> changing direction Same kinematics equations Always separate components! i.e. separate equations for all x quantities, and for all y quantities

Projectile motion Acceleration (in y direction) due to gravity 𝑎 𝑦 =−𝑔 Constant velocity in x direction ( 𝑎 𝑥 =0) Parabolic trajectory Knowns: 𝑎 𝑦 , 𝑎 𝑥 , 𝑣 𝑥 = constant Possible Unknowns: 𝑣 𝑦,𝑓 , 𝑣 𝑦,0 ,∆𝑦,∆𝑥,𝑡, 𝑣 𝑥 (numerical value)

Kinematics Equations for Projectile Motion Constant acceleration in y direction 𝑥 𝑓 = 𝑥 0 + 𝑣 𝑥,0 𝑡+ 1 2 𝑎 𝑥 𝑡 2 = 𝑥 0 + 𝑣 𝑥,0 𝑡 alternatively, ∆𝑥= 𝑣 𝑥,0 𝑡+ 1 2 𝑎 𝑥 𝑡 2 = 𝑣 𝑥,0 𝑡 𝑦 𝑓 = 𝑦 0 + 𝑣 𝑦,0 𝑡+ 1 2 𝑎 𝑦 𝑡 2 alternatively, ∆𝑦= 𝑣 𝑦,0 𝑡+ 1 2 𝑎 𝑦 𝑡 2 = 𝑣 𝑦,0 𝑡− 1 2 𝑔 𝑡 2 𝑣 𝑥,𝑓 = 𝑣 𝑥,0 + 𝑎 𝑥 𝑡= 𝑣 𝑥,0 𝑣 𝑦,𝑓 = 𝑣 𝑦,0 + 𝑎 𝑦 𝑡= 𝑣 𝑦,0 −𝑔𝑡 𝑣 𝑥,𝑓 2 = 𝑣 𝑥,0 2 −2 𝑎 𝑥 ∆𝑥= 𝑣 𝑥,0 2 −2 𝑎 𝑥 𝑥 𝑓 − 𝑥 0 = 𝑣 𝑥,0 2 𝑣 𝑦,𝑓 2 = 𝑣 𝑦,0 2 −2 𝑎 𝑦 ∆𝑦= 𝑣 𝑦,0 2 −2 𝑎 𝑦 𝑦 𝑓 − 𝑦 0 = 𝑣 𝑦,0 2 +2𝑔∆𝑦 Note: when 𝑡 𝑖 ≠0, then 𝑡 above should be ∆𝑡= 𝑡 𝑓 − 𝑡 𝑖

Simplified version Y direction ( 𝑎 𝑦 =−𝑔) X direction 𝑦 𝑓 = 𝑦 0 + 𝑣 𝑦,0 𝑡+ 1 2 𝑎 𝑦 𝑡 2 = 𝑦 0 + 𝑣 𝑦,0 𝑡− 1 2 𝑔 𝑡 2 𝑣 𝑦,𝑓 = 𝑣 𝑦,0 + 𝑎 𝑦 𝑡= 𝑣 𝑦,0 −𝑔𝑡 𝑣 𝑦,𝑓 2 = 𝑣 𝑦,0 2 −2 𝑎 𝑦 ∆𝑦= 𝑣 𝑦,0 2 −2 𝑎 𝑦 𝑦 𝑓 − 𝑦 0 = 𝑣 𝑦,0 2 +2𝑔∆𝑦 X direction 𝑥 𝑓 = 𝑥 0 + 𝑣 𝑥,0 𝑡 𝑣 𝑥,𝑓 = 𝑣 𝑥,0 = 𝑣 𝑥 𝑎 𝑥 =0

Vector Problem A sailboat race course consists of four legs, defined by the displacement vectors A, B, C and D as shown in the drawing. The magnitudes are: A = 2.50 km, B = 5.10 km, and C = 4.50 km. The finish line is 1 km south and 1 km west of the start. (a) Using the data in the drawing, find the distance of the fourth leg (D) and the angle 2. (b) What is the total distance of the course? (c) What is the net displacement of the course?

Practice Problems An object experiences a constant acceleration of 2.00 m/s2 along the −x axis for 2.70 s, attaining a velocity of 16.0 m/s in a direction 45° from the +x axis. Calculate the initial velocity vector of the object. An object is undergoing parabolic motion as shown from the side in Figure 3.42. Assume the object starts its motion at ground level. For the five positions shown, draw to scale vectors representing the magnitudes of (a) the x components of the velocity, (b) the y components of the velocity, and (c) the accelerations.  (Neglect (Neglect any effects due to air resistance.)

Practice Problems An object undergoing parabolic motion travels 100 m in the horizontal direction before returning to its initial height. If the object is thrown initially at a 30° angle from the horizontal, determine the x component and the y component of the initial velocity. (Neglect any effects due to air resistance.) A tiger leaps horizontally out of a tree that is 4.00 m high. If he lands 5.00 m from the base of the tree, calculate his initial speed.(Neglect any effects due to air resistance.)

Practice Problems A boy runs straight off the end of a diving platform at a speed of 5.00 m/s. The platform is 10.0 m above the surface of the water. (a) Calculate the boy’s speed when he hits the water. (b) How much time is required for the boy to reach the water? (c) How far horizontally will the boy travel before he hits the water? (Neglect any effects due to air resistance.) An airplane flying upward at 35.3 m/s and an angle of 30.0° relative to the horizontal releases a ball when it is 255 m above the ground. Calculate (a) the time it takes the ball to hit the ground, (b) the maximum height of the ball, and (c) the horizontal distance the ball travels from the release point to the ground. (Neglect any effects due to air resistance.)

Practice Problem Answers Vector Problem: D = 6.18 km, θ = 33.54 deg., distance = 18.28 km, displacement = 1.414 km 45 deg. W of S P1.1: vx = 16.7 m/s, vy = 11.3 m/s P2.1: vx = 29.13 m/s, vy = 16.82 m/s P2.2: v0 = 5.53 m/s P3.1: vf = 14.87 m/s, t = 1.43 s, Δx = 7.143 m P3.2: t = 9.24 s, Δy = 270.9 m, Δx = 282.4 m