Projectile Motion Horizontal Angular

Rules for Projectile Motion Treat horizontal and vertical as two separate sides of the problems TIME is the key, and the only variable that can be used for both horizontal and vertical Horizontal Motion is always constant vx is constant ax = 0 m/s2 Objects follow a parabolic shape ay = g = -9.81 m/s2

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

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Angular Projectile Motion Animation from www.physicsclassroom.com

Angular Projectile Motion Initial velocity is both vertical and horizontal Use trig functions to find vyi and vx vx = vi Cos θ vyi = vi Sin θ 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/s2 ∆y = 0 m vyf = -vyi ∆x = ?

vi = 55.0 m/s ay = -9.81 m/s2 ∆x = ? θ = 50.0° ∆y = 0.0 m vyf = - vyi - + - vi = 55.0 m/s ay = -9.81 m/s2 ∆x = ? θ = 50.0° ∆y = 0.0 m vyf = - vyi Find vx and vyi vx = vi Cos θ = 55.0 m/s Cos (50.0°) = 55.0 m/s (0.643) = 35.4 m/s vyi = vi Sin θ = 55.0 m/s Sin (50.0°) = 55.0 m/s (0.766) = 42.1 m/s

What do we need to find ∆x? Time! Find time from the vertical side - + - vi = 55.0 m/s ay = -9.81 m/s2 ∆x = ? θ = 50.0° ∆y = 0.0 m vx = 35.4 m/s vyf = - 42.1 m/s vyi = 42.1 m/s What do we need to find ∆x? Time! Find time from the vertical side ∆y = vyi ∆t + ½ ay ∆t2 0 m = (42.1 m/s) ∆t + ½ (-9.81 m/s2) ∆t2 - (42.1 m/s) ∆t = ½ (-9.81 m/s2) ∆t2 (42.1 m/s) = ½ (9.81 m/s2) ∆t ∆t = 8.58 s

Now we can find ∆x ∆x = vx ∆t = (35.4 m/s)(8.58 s) = 304 m - + - + - vi = 55.0 m/s ay = -9.81 m/s2 ∆x = ? θ = 50.0° ∆y = 0.0 m vx = 35.4 m/s ∆t = 8.58 s vyf = - 42.1 m/s vyi = 42.1 m/s Now we can find ∆x ∆x = vx ∆t = (35.4 m/s)(8.58 s) = 304 m

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