9-10-ProjectileMotionII

9-10-ProjectileMotionII - Projectile Motion (General Launch...

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Projectile Motion (General Launch Angle) t x x 0x 0 v + = 2 0 2 1 t a t y y y + + = 0y v H The trajectory is a parabola a y = - 9.8 m/s 2 y 0 v Projectile Motion t v x v y x x 0 v v = t a y y y + = 0 v v x 0 v x 0 v x 0 v x 0 v x 0 v y 0 v y 0 v - 0 v
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Example H s m s m v v o x / 32 . 15 40 cos ) / 20 ( cos 0 0 = = = θ s m s m v v o y / 86 . 12 40 sin ) / 20 ( sin 0 0 = = = To find Maximum height, go to y motion. v 0y = 12.86m/s At maximum height: v y = 0 a y = - 9.8m/s 2, Find y=? ) ( 2 0 2 0 2 y y a v v y y y = initial position x 0 = 0, y 0 = 0 y ) 8 . 9 ( 2 ) 86 . 12 ( 0 2 2 = m y 44 . 8 = maximum height H = 8.44m y v 0 =0 y v =0 y v 2. How much time does it take the ball to reach the maximum height? 1. Find the maximum height H the ball attains. A football is kicked at an angle of θ =40 o above the horizontal axis. Its initial speed is 20 m/s. Ignoring air resistance t ) 8 . 9 ( 86 . 12 0 s t 31 . 1 = t a v v y y y + = 0 Example (continued) 3. How long does the ball spend in the air? H
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9-10-ProjectileMotionII - Projectile Motion (General Launch...

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