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2-dimentional motion

2-dimentional motion. Parabolic or Projectile Motion. Projectile Motion. A projectile is an object moving in two dimensions under the influence of Earth's gravity ; its path is a parabola. Motion in Two Dimensions. Motion in Two Dimensions. a x = 0. a y = - g. Motion in Two Dimensions.

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2-dimentional motion

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  1. 2-dimentional motion Parabolic or Projectile Motion

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

  3. Motion in Two Dimensions

  4. Motion in Two Dimensions ax= 0 ay= -g

  5. Motion in Two Dimensions Ignoring air resistance, the horizontal component of a projectile's acceleration (A) is zero. (B) remains a non-zero constant. (C) continuously increases. (D) continuously decreases.

  6. Solving Problems Involving Projectile Motion • Read the problem carefully, and choose the object(s) you are going to analyze. • Draw a diagram. • Choose an origin and a coordinate system. • Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. • Examine the x and y motions separately.

  7. Solving Problems Involving Projectile Motion 6. List known and unknown quantities. Remember that vx never changes, and 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.

  8. Problem v y d A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. a) How high was the cliff? b) How far from its base did the diver hit the water?

  9. A marble rolls off the edge of a table with a height of 0.755 m and strikes the floor at a distance of 24.3 cm from the edge of the table. Calculate the initial velocity of the marble. • First convert cm to m, 24.3 cm = .243 m ax = 0 dx = 0.243m (how far the ball traveled) viy = 0 g = - 9.80m/s2 dy = - 0.755 (a negative direction) Find vix g = - 9.81 m/s2 - 0.755 m 0.243 m

  10. An archer stands on the wall of a castle and fires an arrow from a height of 12.10 m above the ground. If the archer fires an arrow parallel to the ground with an initial horizontal velocity of 11.0 m/s, how far will the arrow travel horizontally before hitting the ground? ax = 0 vfx = 0 vix = 11.0 m/s viy = 0 g = - 9.80m/s2 dy = - 12.10 (a negative direction) Find dx g = - 9.81 m/s2 - 12.10 m

  11. First find how long the arrow will be in the air dy = 1/2gDt2 Dt = 2dy = 2 (-12.10 m) = 1.57 s -9.81 m/s2 Next, we can make use of the initial velocity in the horizontal and the change in time to find the displacement vix = dx Dt dx= vixDt = (11.0 m/s)(1.57 s) = 17.3 m

  12. An airplane is flying a practice bombing run by dropping bombs on an old shed. The plane is flying horizontally with a speed of 185 m/s. It releases a bomb when it is 593 m away from the shed, and it scores a direct hit. How high was the plane flying when it dropped the bomb? ax = 0 vfx = 0 vix = 185 m/s dx = 593m viy = 0 g = - 9.81m/s2 Find dy 593 m

  13. First we start by using the initial horizontal velocity and the horizontal displacement to determine how long the bomb was in the air vix = dxdx= vixDtDt = dx 593m = 3.21 s Dtvix185 m/s Now that we know how long it takes the bomb to fall, we can calculate the bomb’s vertical displacement. dy = 1/2gDt2 = (0.5)(- 9.81m/s2)(3.21 s)2= - 50.5 m (a negative direction)

  14. A car launches horizontally off the edge of a cliff with a height of 18.6 m. It strikes the ground below at a distance of 98.4 m away from the edge of the cliff. How fast was the car going when it flew off the edge of the cliff? ax = 0 vfx = 0 dx = 98.4 m viy = 0 g = - 9.8 m/s2 dy = - 18.6 m (a negative direction) Find vix - 18.6 m 98.4 m

  15. A car launches horizontally off the edge of a cliff with a height of 18.6 m. It strikes the ground below at a distance of 98.4 m away from the edge of the cliff. How fast was the car going when it flew off the edge of the cliff? ax = 0 vfx = 0 dx = 98.4 m viy = 0 g = - 9.8 m/s2 dy = - 18.6 m (a negative direction) Find vix - 18.6 m 98.4 m

  16. First we start by using the vertical displacement and the acceleration due to gravity to determine how long the car was in the air dy = 1/2gDt2 Dt = 2dy = 2(-18.6 m) = 1.95s g -9.81 m/s Now that we know how long it takes the car to fall, we can calculate the car’s initial velocity. Vix = 98.4 m = 50.5 m/s 1.95 s

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