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Projectile Motion • The Horizontal Motion • The Vertical Motion • The Trajectory • The Horizontal Range • Example • Homework 1 The Horizontal Motion ax = 0 ⇒ vxf = vxi = vi cos θi = constant xf = xi + vxit = xi + (vi cos θi) t 2 (1) The Vertical Motion ay = −g 1 1 yf = yi + vyit − gt2 = yi + (vi sin θi) t − gt2 2 2 vyf = vyi − gt = vi sin θi − gt (2) (3) 2 2 vyf = vyi − 2g (yf − yi) = (vi sin θi)2 − 2g (yf − yi) (4) 3 The Trajectory • • Solving eqn. (1) for t we have xf − x i t= vi cos θi Substituting this into eqn. (2) and setting xi = yi = 0,we get 2 1 x x f f − g yf = (vi sin θi) vi cos θi 2 vi cos θi yf = (tan θi) xf − • g 2 x 2 (vi cos θi)2 f (5) Eqn. (5) has the form y = ax + bx2 where a and b are constants ⇒ the trajectory of a projectile is parabolic 4 The Horizontal Range xf − x i = R Eqn. (1) Eqn. (2) y f − yi = 0 R = (vi cos θi) t 1 ⇒ 0 = (vi sin θi) t − gt2 2 2vi sin θi t= g ⇒ 5 The Horizontal Range (cont’d) 2vi2 R= sin θi cos θi g sin 2θi = 2 sin θi cos θi vi2 R = sin 2θi g Note: R is a maximum when sin 2θi = 1 ⇒ 2θi = 90◦ or θi = 45◦ 6 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? 7 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 = y i 1 2 yf − yi = (vi sin θi) t − gt = 0 2 2vi sin θi 2 (15 m/s) sin 38◦ t= = = 1.9 s g 9.8 m/s2 (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 2 = (vi sin θi)2 − 2g (yf − yi) = 0 vyf vi2 sin2 θi (15 m/s) sin2 38◦ = = 4.4 m yf − y i = 2g 2 (9.8 m/s2) 8 Homework Set 7 - Due Fri. Sept. 24 • Read Section 3.3 • Do Problems 3.9, 3.11 & 3.14 9