Physics I

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Physics I

• Accelerated motion

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Old Business-relative velocity
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Old Business-1-D uniformly accelerated motion

• The Problem Hecht 2-103
• A bag of sand dropped by a would-be assassin from
  the roof of a building just misses Tough-Tony, a
  gangster 2-m tall. The missile traverses the
  height of Tough Tony in 0.20 seconds, landing
  with a thud at his feet. How high was the
  building? Ignore friction.

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Translation

• A particle falling freely (so we know
  acceleration as well) traverses a known distance
  in a known time but without knowing the initial
  or final velocities. Determine the height from
  which the particle fell.

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Given

• ?l 2 meters
• ?t 1/5 second
• aT a g 9.81 m / s2 downward
• No air resistance

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Find

• H height from which particle fell with initial
  speed 0 (it was dropped).

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Problem Type

• Mechanics/Kinematics/Uniform Acceleration/Freely
  Falling particle

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Procedure

• Apply the equations for uniformly accelerated
  motion, knowing the acceleration has a magnitude
  of 9.81 m / s2 and is directed downward

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Calculation

• Choose downward as positive direction.
• For freely falling motion, over the interval of
  Tough Tonys head to feet,
• (vf vi) / t g the acceleration is uniform
• Vav (vf vi)/2 ?x / ?t 2 m/ (1/5 s) 10
  m/s
• (vf vi) 20 m / s
• (vf vi) 9.81m /s2 x (1/5s) 1.962 m / s

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calculations

• Add these two equations to get speed at the
  ground (just before particle thud
• vf 21.962 / 2 m / s 10.981 m / s
• Now we ask, from what height H must the particle
  fall from rest (vi 0) to attain a speed of
  10.985 m / s?
• vf2 2gH
• H vf2 / (2g)
• You do the math!

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New Business-ballistic motion

• In free-fall the horizontal and vertical
  components of the motion are independent
• Non-intuitive
• horizontal rifle over flat earth
• monkey and poison dart

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accelerations

• Vertical
• aT a g downward
• Horizontal
• aT a 0
• Horizontal component of the speed does not change!

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components

• Initial velocity
• Muzzle speed vi
• Angle of elevation ?
• Initial Horizontal component (and does not
  change)
• vix vi cos (?)
• Vertical component
• Viy vi sin(?)

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Special relations

• Time to reach maximum height
• tp vi sin(?) / g
• Time to reach initial level
• tT 2tp 2vi sin(?) / g
• Peak height
• sp vi2 sin2(?)/ (2g)
• Range (along horizontal plane)
• sR vi2 sin(2 ?)/g
• Max range when ? 45 degrees

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Trajectory equation

• sy sx tan(?) (g/2)sx2 /(vix2)

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The Big Gun Game

• The cannon and target

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Examples

• 3-104 A shoe is flung into the air such that at
  the end of 2.0 s it is at its maximum altitude,
  moving at 6.0 m / s. How far away from the
  thrower will it be when it returns to the height
  from which it was tossed?

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Interpreting phrases

• At maximum altitude vy 0 but vx is the same as
  vix
• So here we know that vx vix 6.0 m / s
• At maximum altitude half the elapsed time needed
  to reach launch level is reached.
• tp 2.0 s so tT 2 x 2.0 4.0 s
• How far away the thrower isthats the range
• So sR vix x tT 24 m

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Example 3-124

• A baseball is hit as it comes in, 1.30 m over the
  plate. The blast sends it off at an angle of 30
  degrees above the horizontal with a speed of 45.0
  m/s . The outfield fence is 100 m away and 11.3 m
  high. Ignoring aerodynamic effects, will the ball
  clear the fence?

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Givens

• height of hit wrt ground 1.30 m
• Muzzle velocity of hit ball
• Vi 45.0 m / s
• T 30 degrees above horizontal
• sx 100 m (is or isnt this the range sR too?
• Top of fence from ground 11.3 m.
• aT a g 9.81 m / s2 downward

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Find

• Find sy when sx 100 m
• Does sy put the ball above the fence at that
  distance?

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Problem Type

• Mechanics/kinematics/uniformly accelerated
  motion/ballistic motion

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Procedure

• Translate all lengths to origin at place where
  ball leaves bat.
• Speeds and angles not changed.
• sx not changed
• But top of fence is 11.3m 1.30 m 10.0 m
• Apply the trajectory equation.

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Calculation

• sy sx tan(?) (g/2)sx2 /(vicos(?))2
• Is sy greater than 10.0 m?
• If yes, over fence if no, hits fence.

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Alternative method

• Calculate time for ball to move 100 m
  horizontally
• t sx / (vi cos(?))
• Compute sy from
• sy vi sin(?) t (g/2) t2