Projectile Motion Horizontal Angular Rules for Projectile Motion

Projectile Motion Horizontal Angular

Rules for Projectile Motion Treat horizontal and vertical as two separate sides of the problems l TIME is the key, and the only variable that can be used for both horizontal and vertical l Horizontal Motion is always constant l • vx is constant • ax = 0 m/s 2 Objects follow a parabolic shape l ay = g = -9. 81 m/s 2 l

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Horizontal Projectile Motion l All of the initial velocity is in the x direction, vyi = 0 m/s l Vertical displacement and velocity will always be negative

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Angular Projectile Motion Initial velocity is both vertical and horizontal l Use trig functions to find vyi and vx l • vx = vi Cos θ • vyi = vi Sin θ l Remember clues • vy at the top is 0 m/s • vy at any height is the same while going up and coming down except for direction • ∆y = 0 m if ending at the same height as it started

Example Problem Happy Gilmore hits his shot at 55. 0 m/s with an angle of 50. 0° to the ground. How far did the ball travel before it lands? • vi = 55. 0 m/s • θ = 50. 0° • ay = -9. 81 m/s 2 • ∆y = 0 m • vyf = -vyi • ∆x = ?

- + vi = 55. 0 m/s θ = 50. 0° - l Find vx lvx and vyi = vi Cos θ = 55. 0 m/s Cos (50. 0°) = 55. 0 m/s (0. 643) = 35. 4 m/s lvyi = vi Sin θ = 55. 0 m/s Sin (50. 0°) = 55. 0 m/s (0. 766) = 42. 1 m/s ay = -9. 81 m/s 2 ∆y = 0. 0 m vyf = - vyi ∆x = ?

- + - vi = 55. 0 m/s θ = 50. 0° ay = -9. 81 m/s 2 ∆x = ? ∆y = 0. 0 m vx = 35. 4 m/s vyf = - 42. 1 m/s vyi = 42. 1 m/s l What do we need to find ∆x? l Time! Find time from the vertical side l ∆y = vyi ∆t + ½ ay ∆t 2 0 m = (42. 1 m/s) ∆t + ½ (-9. 81 m/s 2) ∆t 2 - (42. 1 m/s) ∆t = ½ (-9. 81 m/s 2) ∆t 2 (42. 1 m/s) = ½ (9. 81 m/s 2) ∆t ∆t = 8. 58 s

- + vi = 55. 0 m/s θ = 50. 0° ∆t = 8. 58 s - l Now l ay = -9. 81 m/s 2 ∆x = ? ∆y = 0. 0 m vx = 35. 4 m/s vyf = - 42. 1 m/s vyi = 42. 1 m/s we can find ∆x ∆x = vx ∆t = (35. 4 m/s)(8. 58 s) = 304 m

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