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Section 3-5: Projectile Motion

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1 Section 3-5: Projectile Motion

2 A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola. Figure Caption: This strobe photograph of a ball making a series of bounces shows the characteristic “parabolic” path of projectile motion.

3 Projectile Motion Projectile Motion  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/s2 Goal: Describe 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

4 Projectile Motion  vy = gt, y = (½)g t2  vx = vx0 , x = vx0t
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: Down is positive & the origin is at table top: y0 = 0. Initially, there is no y component of velocity: vy0 = 0  vy = gt, y = (½)g t2 x part of motion: The origin is at the table top: x0 = 0. No x component of acceleration(!): a = 0. Initially the x component of velocity is: vx0  vx = vx0 , x = vx0t

5 Ball Rolls Across Table & Falls Off
t = 0 here Can be understood by analyzing horizontal vertical motions separately. Take down as positive. Initial velocity has an x component ONLY! That is vy0 = 0. Figure Caption: Projectile motion of a small ball projected horizontally. The dashed black line represents the path of the object. The velocity vector at each point is in the direction of motion and thus is tangent to the path. The velocity vectors are green arrows, and velocity components are dashed. (A vertically falling object starting at the same point is shown at the left for comparison; vy is the same for the falling object and the projectile.) At any point, v has both x & y components. Kinematic equations tell us that, at time t, vx = vx0, vy = gt x = vx0t y = vy0t + (½)gt2

6 Summary: Ball rolling across table & falling.
Vector velocity v has 2 components: vx = vx0 , vy = gt Vector displacement D has 2 components: x = vx0t , y = (½)g t2

7 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 one 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. Figure Caption: Multiple-exposure photograph showing positions of two balls at equal time intervals. One ball was dropped from rest at the same time the other was projected horizontally outward. The vertical position of each ball is seen to be the same at each instant.

8 Projectile Motion PHYSICS: y part of motion: vy = gt , y = (½)g t2
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)

9 General Case: Object is launched at initial angle θ0 with the horizontal. Analysis is similar to before, except the initial velocity has a vertical component vy0  0. Let up be positive now! vx0 = v0cosθ0 vy0 = v0sinθ0 but, acceleration = g downward for the entire motion! Figure Caption: Path of a projectile fired with initial velocity v0 at angle θ0 to the horizontal. Path is shown dashed in black, the velocity vectors are green arrows, and velocity components are dashed. The acceleration a = dv/dt is downward. That is, a = g = -gj where j is the unit vector in the positive y direction. LLLL Parabolic shape of path is real (neglecting air resistance!)

10 General Case: Take y positive upward & origin at the point where it is shot: x0 = y0 = 0
vx0 = v0cosθ0, vy0 = v0sinθ0 Horizontal motion: NO ACCELERATION IN THE x DIRECTION! vx = vx0 , x = vx0 t Vertical motion: vy = vy0 - gt , y = vy0 t - (½)g t2 (vy) 2 = (vy0)2 - 2gy If y is positive downward, the - signs become + signs. ax = 0, ay = -g = -9.8 m/s2

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

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

13 Example 3-4: 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, 90.0 m from the base of the cliff where the cameras are? y is positive upward, y0 = 0 at top. Also vy0 = 0 vx = vx0 = ? vy = -gt x = vx0t, y = - (½)gt2 Time to Bottom: t = √2y/(-g) = 3.19 s vx0 = (x/t) = 28.2 m/s

14 Example 3-5: Kicked football
A football is kicked at an angle θ0 = 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. θ0 = 37º, v0 = 20 m/s  vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s lllllllll

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