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Parabolic Motion

Parabolic Motion. Movement in two dimensions. Shape of the Motion. The motion is parabolic in shape:. Forces acting on the projectile:. In the y-direction, we have gravity g. Gravity is pointing downwards towards the ground In the x-direction, we have no forces acting upon the object.

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Parabolic Motion

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  1. Parabolic Motion Movement in two dimensions

  2. Shape of the Motion The motion is parabolic in shape:

  3. Forces acting on the projectile: • In the y-direction, we have gravity g. • Gravity is pointing downwards towards the ground • In the x-direction, we have no forces acting upon the object. • It will remain in constant motion • In the x-direction, the velocity will be vx=v cos() • In the y-direction, the velocity will be vy=v sin() -g ∆t

  4. Horizontal motion Uniform motion in the form of: v = Δd / ∆t

  5. Vertical motionORThe five equations of accelerated motion (I) a = vf– vi / Δt (II) Δd = vi Δt + 1/2 a (Δt)2 (III) Δd = 1/2 (vf+ vi) Δt (IV) vf2= vi2+ 2 a Δd (V) Δd = vfΔt – 1/2 a(Δt)2

  6. Example 1 • Wile E. Coyote is running after road-runner but fails to catch him and runs off of the edge of a 100m-high cliff with a velocity of 45 m/s. How far does he get from the edge of the cliff?

  7. What Happens in Example 1 • Let's say the coyote does run off of the cliff. As he leaves the cliff he has a horizontal velocity. • Since we are ignoring air resistance (which is a very good idea since it will be practically zero), then there is no horizontal force to cause an horizontal acceleration. • Since there is no horizontal acceleration, the coyote will travel horizontally at the same speed the whole time! • As soon as the coyote leaves the cliff he will experience a vertical force due to gravity. • This force will cause him to start to accelerate in the vertical direction. • As he falls he will be going faster and faster in the vertical direction

  8. Example 1, Solved • First we analyze how long Wile E. Coyote is in the air, and, to do so, we look at the y-component: • Δd = vi Δt + 1/2 a (Δt)2 100=0 Δt + 1/2 9.8 (Δt) 2Δt=3.19 s • Then we look at the x-component to see how far he got: • v = Δd / ∆t 45=Δd / 3.19 Δd=143.55

  9. Example 2 A basketball player shoots a basketball from his chest at an initial height of 1 m with an initial upward velocity of 10 m/s. What is the height of the basketball after 2 seconds? Share with a partner and discuss how to solve it.

  10. Example 2, solved • Use the equation: Δd = vi Δt + 1/2 a (Δt)2Δd = 10(2) + 1/2 (9.8)(22) • Δd = 1.4 m

  11. Let’s work together A stone is thrown from the top of a building upward at an angle 30.0° to the horizontal and with an initial speed of 20.0 m/s. If the height of the building is 45.0 m: How long is it before the stone hits the ground? What is the speed of the stone before it strikes the ground? Where does the stone strike the ground?

  12. Maximum Height • The maximum height of an object in projectile motion is at the peak of the parabola. • The vertical velocity at that point is v=0. • We can use a = vf– vi/ Δt • Therefore, the time it takes to reach that point is Δt= vi /g • Since h = vi Δt + 1/2 a (Δt)2 • Thus, the maximum height it reaches is h=v2/2g

  13. Questions? Comments or concerns?

  14. Exit card (insert marks here) A 2kg object is thrown from a height of with a velocity of 15 m/s at an angle of 50, how long will it take for it to hit the ground? A 2000 kg car drives off of a ramp of a height of 10 m. At what distance will a ramp of 2m need to be placed so that the car can land safely? A golf ball is hit with a velocity of 50 m/s at an angle of 60. What is its maximum height?

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