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Projectile Motion - Horizontal launches

Projectile Motion - Horizontal launches. The scale on the left is 1cm = 5m Draw the positions of the dropped ball every second that it is in the air. Neglect air resistance and assume g = 10 m/s 2 . Estimate the number of seconds that the ball is in the air.

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Projectile Motion - Horizontal launches

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  1. Projectile Motion - Horizontal launches The scale on the left is 1cm = 5m Draw the positions of the dropped ball every second that it is in the air. Neglect air resistance and assume g = 10 m/s2. Estimate the number of seconds that the ball is in the air Four positions of a ball thrown horizontally without gravity are shown. Draw the positions of the ball with gravity. Describe the path taken by the ball. It is parabolic How is motion in the vertical direction affected by motion in the horizontal direction? They are independent of each other

  2. Projectile Motion Rules: 1. Vertical Motion is effected by gravity 2. Horizontal motion is constant Motion in the horizontal direction is independent of the motion in the vertical direction. 4. Assume no air resistance

  3. Physical Expressions Horizontal Vertical vHi = vi cos vvi = vi sin vvf = vVi + g t vHi = vHf = vH dv = vvi t + 1/2 g t2 dH = vH t a = g = -9.8 m/s2 a = 0 m/s2

  4. Example Problem A projectile is launched from a height of 44.1 m with a initial horizontal speed of 20 m/s. a) How long is it the air? b) how far does it travel horizontally before it hits the ground? vvi = 0 m/s vH = 20 m/s dv = - 44.1 m g = - 9.8 m/s2 a) dv = vvi t + 1/2 g t2 20m/s - 44.1 m= (0 m/s) t + 1/2 (-9.8 m/s2) t2 - 44.1 m t =  (2 (-44.1 m) / -9.8 m/s2) = 3 s b) dH = vH t dH = (20 m/s) (3s) = 60m

  5. Example Continued vH  c) What velocity does the projectile hit the ground? vvf v vvi = 0 m/s vH = 20 m/s dv = -44.1 m g = - 9.8 m/s2t = 3s vVf = vVi + g t vVf = (0 m/s) + (-9.8 m/s2) (3s) = - 29.4 m/s v =  (vvf2 + vH2) = (- 29.4 m/s)2 + (20m/s)2 = 35.6 m/s  = tan-1 (vvf / vH) below the horizontal  = tan-1 (29.4 m/s / 20 m/s) = 55.8 0 below the horizontal v = 36 m/s @ 560 below the horizontal

  6. Example Continued vH  or… can solve for speed using conservation of energy vvf vBottom vTop= 20 m/s hTop = 44.1 m hBottom = 0m g = -9.8 m/s2 vBottom = ? KETop + GPETop = KEBottom + GPEBottom 1/2 m vTop2 + (-mghTop) = 1/2 m vBottom2 + (-mghBottom) 1/2 vTop2 + (-ghTop) = 1/2 vBottom2 + (-ghBottom) 1/2 (20 m/s)2 + (-(-9.8 m/s2) (44.1m) = 1/2 vBottom2 + (-(-9.8m/s2)(0m) 632.2 m2/s2 = 1/2 vBottom2 vBottom =  2(632.2 m2/s2) = 35.6 m/s

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