Gravity Physics Example 3

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Example 3

easy
A ball is dropped from rest near Earth's surface. How fast is it going after 3 seconds3 \text{ seconds}? Ignore air resistance and use g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution

  1. 1
    Gravity provides a constant downward acceleration of g=9.8 m/s2g = 9.8 \text{ m/s}^2. The ball is dropped, so initial velocity v0=0v_0 = 0.
  2. 2
    Apply the kinematic equation for velocity: v=v0+gtv = v_0 + gt.
  3. 3
    Substitute values: v=0+9.8×3=29.4 m/s downwardv = 0 + 9.8 \times 3 = 29.4 \text{ m/s downward}

Answer

v=29.4 m/s downwardv = 29.4 \text{ m/s downward}
Near Earth's surface, gravity accelerates all objects equally at g≈9.8 m/s2g \approx 9.8 \text{ m/s}^2 regardless of mass, assuming no air resistance.

About Gravity

The universal attractive force between any two objects with mass, decreasing with the square of distance.

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