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Section 3-5: Projectile Motion

Section 3-5: Projectile Motion. 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  Motion of an object that is projected into the air at an angle.

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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.

  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 • 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. 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.

  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 θ0with 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! 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. • Listknown & 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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