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Projectile Motion Horizontal Angular.

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Presentation on theme: "Projectile Motion Horizontal Angular."— Presentation transcript:

1 Projectile Motion Horizontal Angular

2 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 = m/s2

3 Horizontal Projectile Motion
Animation from

4 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

5 Interesting Applications
Animation from

6 Angular Projectile Motion
Animation from

7 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

8 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 = m/s2 ∆y = 0 m vyf = -vyi ∆x = ?

9 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 = 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

10 What do we need to find ∆x? Time! Find time from the vertical side
- vi = 55.0 m/s ay = m/s2 ∆x = ? θ = 50.0° ∆y = 0.0 m vx = 35.4 m/s vyf = 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

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

12 Where Should We Aim the Cannon?
At or Above the monkey? Animation from

13 Above the Monkey Animation from

14 At the Monkey Animation from

15 At the Monkey (Faster) Animation from


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