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Projectile Motion: Understanding the Path and Velocity of Projectiles

Learn about the shape of the projectile path, known as a parabola, and how to calculate the distance and time of a projectile's motion. Understand the independent vertical and horizontal motions of projectiles and their components.

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Projectile Motion: Understanding the Path and Velocity of Projectiles

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

  2. Projectile Motion • PARABOLA • Can be represented mathematically • Same old stuff

  3. 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!!!!!

  4. 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

  5. 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? X Y

  6. Projectile Problems • Organize information in terms of X and Y components X Y

  7. Projectile Problems • Organize information in terms of X and Y components X Y Vx = 9.0 m/s dx = ? Δdy = 74m

  8. Projectile Problems • What information, that is not stated in the problem, do we know. X Y Vx = 9.0 m/s dx = ? Δdy = 74m ay = 9.8 m/s2 vyi = 0 m/s  at peak of parabola for 2nd half of the trip

  9. Projectile Problems • What one variable is part of both the x and y components? TIME • To solve for dx, we need the time vx = dx / t

  10. Projectile Problems • Can’t solve for t using the X components? • Can we solve for t using the Y? YEP!

  11. Projectile Problems Y’s Δdy = 74m ay = 9.8 m/s2 vyi = 0 m/s  at peak of parabola t = ? Use 2nd half of parabolic motion, where vyi = 0 m/s (peak of parabola) Δdy = vyit + ½ ayt2

  12. Projectile Problems • Solve for t dy = vit + ½ at2 74m = 0 + ½ (9.8m/s2) t2 Therefore, t = √(74m / (½ (9.8m/s2) ) t = 3.9s

  13. Projectile Problems • Knowing the time (3.9s) and vx (9.0m/s), we can solve for dx (where we should look for our keys) • Using Vx = dx / t 9.0m/s = dx / 3.9s dx = 35 m We should look 35 m from the cliff

  14. 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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