Projectile Motion Motion in Two Dimensions A projectile

Projectile Motion

Motion in Two Dimensions A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

Projectile Motion �Projectile Motion of an object that is projected into the air at an angle. �Near the Earth’s surface, the acceleration a on the projectile is downward and equal to a = g = - 9. 8 m/s 2 �Goal: Describe motion after it starts. �Galileo: Analyzed horizontal & vertical components of motion separately. �Remember: Displacement D & velocity v are vectors Components of motion can be treated separately

The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g. Photo shows two balls that start to fall at the same time. The on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

Projectile Motion �Simplest example: A ball rolls across a table, to the edge & falls off the edge to the floor. It leaves the table at time t = 0. Analyze the y part of motion & the x part of motion separately. �y part of motion: Up is positive & the origin is at table top: y 0 = 0. Initially, there is no y component of velocity: vy 0 = 0 vy = - gt, y = - (½)g t 2 �x part of motion: The origin is at the table top: x 0 = 0. No x component of acceleration(!): a = 0. Initially the x component of velocity is: vx 0 vx = vx 0 , x = vx 0 t

Ball Rolls Across Table & Falls Off t = 0 here Can be understood by analyzing horizontal vertical motions separately. Take up as positive. Initial velocity has an x component ONLY! That is vy 0 = 0. At any point, v has both x & y components. Kinematic equations tell us that, at time t, vx = vx 0, vy = - gt x = vx 0 t, y = - (½)gt 2

�Summary: Ball rolling across table & falling. �Vector velocity v has 2 components: vx = vx 0 , vy = - gt �Vector displacement D has 2 components: x = vx 0 t , y = - (½)g t 2

Projectile Motion �PHYSICS: y part of motion: vy = - gt , y = - (½)g t 2 The SAME as free fall motion!! An object projected horizontally will reach the ground at the same time as an object dropped vertically from the same point! (x & y motions are independent)

General Case: Object is launched at initial angle θ 0 with the horizontal. Analysis is similar to before, except the initial velocity has a vertical component vy 0 0. Let up be positive now! vx 0 = v 0 cosθ 0 , vy 0 = v 0 sinθ 0 but, acceleration = g downward for the entire motion! LLLL Parabolic shape of path is real (neglecting air resistance!)

�General Case: Take y positive upward & origin at the point where it is shot: x 0 = y 0 = 0 vx 0 = v 0 cosθ 0, vy 0 = v 0 sinθ 0 �Horizontal motion: NO ACCELERATION IN THE x DIRECTION! vx = vx 0 , x = vx 0 t �Vertical motion: vy = vy 0 - gt , y = vy 0 t - (½)g t 2 (vy) 2 = (vy 0)2 - 2 gy � If y is positive downward, the - signs become + signs. ax = 0, ay = -g = -9. 8 m/s 2

Summary: Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.

Solving Problems Involving Projectile Motion 1. Read the problem carefully, &choose the object(s) you are going to analyze. 2. Sketch a diagram. 3. Choose an origin & a coordinate system. 4. Decide on the time interval; this is the same in both directions, & includes only the time the object is moving with constant acceleration g. 5. Solve for the x and y motions separately. 6. List known & unknown quantities. Remember that vx never changes, & that vy = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

Example: Driving off a cliff!! A movie stunt driver on a motorcycle speeds horizontally off a 50. 0 -m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90 m from the base of the cliff where the cameras are? vx = vx 0 = ? vy = -gt x = vx 0 t, y = - (½)gt 2 Time to Bottom: t = √ 2 y/(-g) = 3. 19 s vx 0 = (x/t) = 28. 2 m/s y is positive upward, y 0 = 0 at top. Also vy 0 = 0

Example: Kicked football �A football is kicked at an angle θ 0 = 37. 0° with a velocity of 20. 0 m/s, as shown. Calculate: a. Max height. b. Time when hits ground. c. Total distance traveled in the x direction. d. Velocity at top. e. Acceleration at top. θ 0 = 37º, v 0 = 20 m/s vx 0= v 0 cos(θ 0) = 16 m/s, vy 0= v 0 sin(θ 0) = 12 m/s