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Projectile motion

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Now Playing:Projectile motion – Example 1a
Examples
  1. Type 1 – Projectile at an Upward Angle
    A projectile is launched from the edge of a cliff 100 m above ground level with an initial speed of 20 m/s at an angle of 35° above the horizontal. Find:
    projectile launched off cliff
    1. the horizontal and vertical components of the launch speed.

    2. the velocity vector at the maximum height.

    3. the maximum height above the cliff top reached by the projectile.

    4. the time taken to reach the maximum height.

    5. the projectile's time of flight.

    6. the projectile's range.

    7. the projectile's impact velocity (magnitude and direction) with the ground.

  2. time of flight: t=2vi⋅sinθg t=\frac{2v_i \cdot sin\theta }{g}
  3. horizontal range: dx=vi2⋅sin2θgd_x = \frac{v_i^2 \cdot sin2\theta}{g}
    Projectile motion
    Jump to:Notes
    Notes
    Notes:

    Shortcut: Level Range Formulas for projectile returning to its original height:
    • horizontal range: dx=vi2⋅sin2θgd_x = \frac{v_i^2 \cdot sin2\theta}{g}
    • time of flight: t=2vi⋅sinθg t=\frac{2v_i \cdot sin\theta }{g}