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Projectile Motion ► All objects move in air along a similar path. Explain the shape of that path. ► What is this curve called? (math class)

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Presentation on theme: "Projectile Motion ► All objects move in air along a similar path. Explain the shape of that path. ► What is this curve called? (math class)"— Presentation transcript:

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3 Projectile Motion ► All objects move in air along a similar path. Explain the shape of that path. ► What is this curve called? (math class)

4 Projectile Motion ► PARABOLA PARABOLA ► Can be represented mathematically ► Same old stuff

5 Projectile Motion Combining the Laws of Motion and what we know about vectors, we can predict the path of projectiles. REMEMBER… X & Y COMPONENTS ARE INDEPENDENT OF EACH OTHER!!!!!

6 Projectile Motion ► Projectile – An object with independent vertical (y) and horizontal (x) motions that moves through the air only under the influence of gravity after an initial thrust ► Trajectory – the path of a projectile through the air

7 Projectile Problem You accidentally throw your car keys horizontally at 9.0 m/s from a cliff 74 m high. How far away from the base of the cliff should you look for your keys? XY

8 Projectile Problems ► Organize information in terms of X and Y components XY

9 Projectile Problems ► Organize information in terms of X and Y components XY Δd y = 74mV x = 9.0 m/s d x = ?

10 Projectile Problems ► What information, that is not stated in the problem, do we know. XY Δd y = 74m a y = 9.8 m/s 2 v yi = 0 m/s  at peak of parabola for 2 nd half of the trip V x = 9.0 m/s d x = ?

11 Projectile Problems ► What one variable is part of both the x and y components? TIME ► To solve for d x, we need the time v x = d x / t

12 Projectile Problems ► Can’t solve for t using the X components? ► Can we solve for t using the Y? YEP!

13 Projectile Problems Y’s Δd y = 74m a y = 9.8 m/s 2 v yi = 0 m/s  at peak of parabola t = ? Use 2 nd half of parabolic motion, where v yi = 0 m/s (peak of parabola) Δd y = v yi t + ½ a y t 2

14 Projectile Problems ► Solve for t d y = v i t + ½ at 2 74m = 0 + ½ (9.8m/s 2 ) t 2 Therefore, t = √(74m / (½ (9.8m/s 2 ) ) t = 3.9s

15 Projectile Problems ► Knowing the time (3.9s) and v x (9.0m/s), we can solve for d x (where we should look for our keys) ► Using V x = d x / t 9.0m/s = d x / 3.9s d x = 35 m d x = 35 m We should look 35 m from the cliff

16 Projectile Motion Practice Problem ► A stone is thrown horizontally at a speed of +5.0 m/s from the top of a cliff 78.4 m high.  How long does it take the stone to reach the bottom of the cliff?  How far from the base of the cliff does the stone strike the ground?  What are the horizontal and vertical components of the velocity of the stone just before it hits the ground?


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