A ball is dropped on the floor from a height of \(20 \text{ m}\). It rebounds to a height of \(5 \text{ m}\). Ball remains in contact with floor for \(1 \text{ s}\). The average acceleration during contact is (\(g = 10 \text{ m/s}^2\))
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Remember that velocity is a vector; since the direction changes, you must add the magnitudes to find the total change in velocity.
Step 1: Understanding the Question:
The average acceleration during impact is the change in velocity divided by the contact time. Velocity changes direction upon rebound, so signs are critical. Step 2: Key Formula or Approach:
1) Velocity just before hitting: \(v_1 = \sqrt{2gh_1}\).
2) Velocity just after rebounding: \(v_2 = \sqrt{2gh_2}\).
3) Average acceleration: \(a = \frac{\vec{v_2} - \vec{v_1}}{\Delta t}\). Step 3: Detailed Explanation:
Let downward direction be negative and upward be positive.
Velocity before impact (downward):
\[ v_1 = -\sqrt{2 \times 10 \times 20} = -\sqrt{400} = -20 \text{ m/s} \]
Velocity after impact (upward):
\[ v_2 = +\sqrt{2 \times 10 \times 5} = +\sqrt{100} = +10 \text{ m/s} \]
Change in velocity:
\[ \Delta v = v_2 - v_1 = 10 - (-20) = 30 \text{ m/s} \]
Average acceleration:
\[ a_{avg} = \frac{\Delta v}{\Delta t} = \frac{30}{1} = 30 \text{ m/s}^2 \] Step 4: Final Answer:
The average acceleration is \(30 \text{ m/s}^2\).