Download Presentation
## Parabolic Motion

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**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. • 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**Horizontal motion**Uniform motion in the form of: v = Δd / ∆t**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**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?**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**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**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.**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**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?**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**Questions?**Comments or concerns?**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?