1.5 projectile motion

1.5 projectile motion

• 1.
  LECTURE 5 Projectile Motion IBPhysics Power Points Topic 9.1 Motion in Fields www.pedagogics.ca
• 2.
  Projectile Motion IsEverywhere
• 3.
  A projectile isan object moving in two dimensions under the influence of only Earth's gravity; its path is a parabola.
• 4.
  Projectile motion canbe understood by analyzing the horizontal and vertical motions separately.
• 5.
  The velocity inthe x-direction is constant; in the y-direction the object has a constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial horizontal velocity. 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.
• 6.
  Key Concept: A projectileis a free falling object that moves sideways. Vertically, a projectile is no different than a dropped object or an object that has been thrown straight upwards or downwards. g = - 9.81 ms-2
• 7.
  Key Concept: A projectileis a free falling object that moves sideways. Horizontally, a projectile is no different than any other object moving with a constant velocity. Gravity has NO effect on the horizontal motion. a = 0 ms-2
• 8.
  A horizontally launchedprojectile will have an initial vertical velocity of zero. It is a dropped object moving sideways. Compare the flight time for each of these three paths:
• 9.
  A diver running1.8 ms-1 dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. How high was the cliff, and how far from its base did the diver hit the water? Δsy = uyΔt + ½gΔt2 = 0 + ½(-9.81)(3.0)2 = - 44 m Δsx = uxΔt = (1.8)(3.0) = 5.4 m
• 10.
  If an objectis launched at an initial angle of θ with the horizontal, the analysis is similar except that the initial velocity now has a vertical component.
• 11.
  A projectile isfired with an initial speed of 65.2 ms-1 at an angle of 34.5°above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after firing.
• 12.
  Solving Problems InvolvingProjectile Motion Projectile motion is motion with constant acceleration in two dimensions. ax= 0 ay = g = -9.81 ms-2 Horizontal motion Vertical motion vx = ux vy = uy+gΔt sx = uxΔt Δsy = uyΔt +½gΔt2 v2 y= u2 y +2g(Δsy )
• 13.
  1. Read theproblem carefully. Draw a diagram. 2. Choose an origin and a coordinate system. 3. 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. 4. Examine the x and y motions separately. 5. List known and unknown quantities. Remember that ux never changes, and that vy = 0 at the highest point.