Projectile Motion A projectile is an object moving

Projectile Motion

• A projectile is an object moving in 2 dimensions under the influence of Earth's gravity; its path is an upside down 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 projectile motion after it starts. • Galileo: Analyzed horizontal & vertical components of motion separately. • Today: Displacement D & velocity v are vectors Components of motion can be treated separately

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 the motion: Down 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: Origin is at the table top: x 0 = 0. There is NO x component of acceleration! ax = 0. Initially the x component of velocity is vx 0 vx = vx 0, x = vx 0 t

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

Summary • A ball rolling across the table & falling. • Vector velocity v has 2 components: vx = vx 0 , vy = gt • Vector displacement D has 2 components: x = vx 0 t , y = 2 (½)gt

• The speed in the x direction is constant; in the y-direction the object moves with constant acceleration g. • The photo shows 2 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 the vertical positions of the 2 balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

Projectile Motion PHYSICS • Vertical (y) part of the motion: vy = gt , y = (½)g 2 t 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)

Projectile Motion A common example of a projectile is a baseball!

A “Somewhat General” Case • An object is launched at initial angle θ 0 with the horizontal. • Analysis of the motion is similar to before, except the initial velocity has a vertical component vy 0 0. Let up be positive now!

A “Somewhat General” Case Let up be positive! Components of initial velocity v 0: vx 0 = v 0 cosθ 0 vy 0 = v 0 sinθ 0 The Parabolic shape of the path is real. Acceleration = g down for the entire trip!

• 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 (vy) 2 = (vy 0)2 - 2 gy 2 t • If y is positive down, the - signs become + signs.

Summary: Projectile Motion with constant acceleration in 2 dimensions, where the acceleration is g and is down.

Solving Projectile Motion Problems 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, & vy = 0 at highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

Projectile Motion Example 4. 4: Rolling off a Cliff • A car rolls off a cliff of height h = 20 m. Its initial velocity is v 0 = 10 m/s, along the horizontal. Calculate a. The time it takes to hit the ground at the base after it leaves the cliff. b. Its horizontal distance x from the base when it hits the ground. c. Its velocity when it hits the ground.

Example: Driving off a cliff!! A movie stunt driver on a motorcycle speeds horizontally off a 50 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 y is positive upward, y 0 = 0 at top. Also vy 0 = 0

Example: Driving off a cliff!! A movie stunt driver on a motorcycle speeds horizontally off a 50 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 lllll • 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