1. Identify the initial conditions:
- Initial height, h = 20 \, \text{m}
- Initial velocity, v_0 = 0 \, \text{m/s} (since the ball is dropped)
- Gravitational acceleration, g = 9.8 \, \text{m/s}^2
2. Use the formula for the distance under constant acceleration to solve for time ( t ):
h = v_0 t + \frac{1}{2} g t^2
3. Plug in the known values:
20 = 0 + \frac{1}{2} \cdot 9.8 \cdot t^2
4. Simplify the equation:
20 = 4.9 t^2
5. Solve for t :
t^2 = \frac{20}{4.9}
6. Calculate the square root:
t = \sqrt{\frac{20}{4.9}}
7. Thus,
t \approx 2 \, \text{s}