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Projectile Motion.

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

1 Projectile Motion

2 Projectiles launched at an angle
Start Stop

3 Projectiles launched at an angle
vyf vy vx vx vy Δy θ vi Start Stop Δx

4 Projectiles launched at an angle
Δx = vi(cos )Δt (horizontal distance) vyf = vi(sin ) – gΔt (final vertical velocity) vyf2= vi2(sin )2 – 2gΔy (final vertical velocity) Δy = vi(sin )Δt – ½ gΔt2 (vertical distance) if you are finding time with either of these two formulas, you are only covering half of the parabola

5 A projectile was launched at an angle of 22
A projectile was launched at an angle of 22.0o with an initial velocity of 25.0 m/s. a. How high will it go?

6 b. How long will it take to reach the top of the flight?

7 c. How long did it take the projectile to land?

8 d. How far down the field did the projectile travel?


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