Projectile Motion The Horizontal Motion The Vertical Motion The - - PDF document

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Projectile Motion The Horizontal Motion The Vertical Motion The - - PDF document

Feb 10, 2024 255 likes •367 views

Projectile Motion The Horizontal Motion The Vertical Motion The Trajectory The Horizontal Range Example Homework 1 The Horizontal Motion a x = 0 v xf = v xi = v i cos i = constant x f = x i + v xi t = x i + ( v i

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SLIDE 1

Projectile Motion

  • The Horizontal Motion
  • The Vertical Motion
  • The Trajectory
  • The Horizontal Range
  • Example
  • Homework

1

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SLIDE 2

The Horizontal Motion

ax = 0 ⇒ vxf = vxi = vi cos θi = constant xf = xi + vxit = xi + (vi cos θi) t (1)

2

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SLIDE 3

The Vertical Motion

ay = −g yf = yi + vyit − 1 2gt2 = yi + (vi sin θi) t − 1 2gt2 (2) vyf = vyi − gt = vi sin θi − gt (3) v2

yf = v2 yi − 2g (yf − yi) = (vi sin θi)2 − 2g (yf − yi) (4)

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SLIDE 4

The Trajectory

  • Solving eqn. (1) for t we have

t = xf − xi vi cos θi

  • Substituting this into eqn. (2) and setting xi = yi =

0,we get yf = (vi sin θi)

  

xf vi cos θi

   − 1

2g

  

xf vi cos θi

  

2

yf = (tan θi) xf −

   

g 2 (vi cos θi)2

    x2

f

(5)

  • Eqn. (5) has the form y = ax + bx2 where a and

b are constants ⇒ the trajectory of a projectile is parabolic

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SLIDE 5

The Horizontal Range

xf − xi = R yf − yi = 0

  • Eqn. (1)

⇒ R = (vi cos θi) t

  • Eqn. (2)

⇒ 0 = (vi sin θi) t − 1 2gt2 t = 2vi sin θi g

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SLIDE 6

The Horizontal Range (cont’d)

R = 2v2

i

g sin θi cos θi sin 2θi = 2 sin θi cos θi R = v2

i

g sin 2θi Note: R is a maximum when sin 2θi = 1 ⇒ 2θi = 90◦ or θi = 45◦

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SLIDE 7

Example

A soccer player kicks a ball at an angle of 38◦ from the horizontal, with an initial speed of 15 m/s. (a) How long is the ball in the air? (b) How far down the field does the ball land? (c) How high does the ball go?

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SLIDE 8

Example Solution

A soccer player kicks a ball at an angle of 38◦ from the horizontal, with an initial speed of 15 m/s. (a) How long is the ball in the air? vi = 15 m/s θi = 38◦ yf = yi yf − yi = (vi sin θi) t − 1 2gt2 = 0 t = 2vi sin θi g = 2 (15 m/s) sin 38◦ 9.8 m/s2 = 1.9 s (b) How far down the field does the ball land? xf−xi = (vi cos θi) t = (15 m/s) (cos 38◦) (1.9 s) = 22 m (c) How high does the ball go? vyf = 0 v2

yf = (vi sin θi)2 − 2g (yf − yi) = 0

yf − yi = v2

i sin2 θi

2g = (15 m/s) sin2 38◦ 2 (9.8 m/s2) = 4.4 m

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SLIDE 9

Homework Set 7 - Due Fri. Sept. 24

  • Read Section 3.3
  • Do Problems 3.9, 3.11 & 3.14

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