Five Chapter 3

• No matter how heavy an object is, when dropped in Earth's gravitational field it accelerates at -9.81 m/s².
• When an object is thrown horizontally when gravity is present, the object has two types of motion:
  • uniform motion horizontally
    • There is no acceleration in the horizontal direction because acceleration due to gravity is perpendicular to this direction. There is no component of the acceleration due to gravity in the horizontal direction. Horizontal velocity cannot change unless there is a component of acceleration parallel to its direction. Remember that we called perpendicular vectors mutually independent (They do not affect each other).
  • uniform acceleration vertically
• Projectile - An object moving freely under the influence of gravity.
• When an object is thrown horizontally in gravity:
  • the vertical displacement traveled in equal time intervals increases downwards.
  • the horizontal distance traveled in equal time intervals remains constant.
  • The path taken by the projectile is a parabola.
• Trajectory - The parabolic path taken by projectiles.
• In general, for projectile motion, vertical motion does not affect the horizontal motion, and vice versa.
• For projectile motion, the instantaneous velocity is in a direction that is tangent to the trajectory.
• The instantaneous velocity can be resolved into a horizontal and a vertical component.
• The horizontal component can be dealt with as uniform motion. The vertical component is dealt with as uniformly accelerated motion.

Here is an example of projectile motion (Two Dimensional Motion)

• A vacant house is used as a target at an army practice range. An artillery gun is aimed at the house at an angle of 30^o to horizontal. The speed of the artillery shell when it leaves the gun (muzzle speed) is 500 m/s. If the house is at the same elevation as the gun, and is 22.1 km away, does the shell hit the house?

• This calculation tells us how far horizontally the shell travels before landing.
• If we solve the above problem by replacing all the numbers with letters, we derive a set of equations that applies to this situation.

• It should be noted that the last two equations enclosed in rectangles can only be used if the shell lands at the same elevation as the artillery gun is at.
• If the shell does not land at that same elevation as the artillery gun, then the more general solution given in the example above must be followed with a minor change. The minor change is that the Dd is no longer zero when solving for the time the shell is in the air. The Dd equals what ever displacement the shell lands at above or below the elevation of the artillery gun. For example, if the house is in a valley 40 m below the artillery gun, then Dd = -40 m. On the other hand, if the house is on a hill 40 m above the artillery gun, then Dd = +40 m.
• For the last formula above, the maximum Dd_h occurs when sin(2q) = 1 or 2q = 90^o or q = 45^o.
• Every range other than the maximum range, has two possible angles that the artillery gun can be aimed at, in order to get the shell to the desired target.

• The equations:

can only be used if

• air resistance can be ignored.
• acceleration due to gravity is constant.
• the Earth is flat.
• the projectile remains in the same vertical plane.

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October 30, 2013