Question

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 (\(g = 10 \text{ m/s}^2\))

• A. \(30 \text{ m/s}^2\)
• B. \(20 \text{ m/s}^2\)
• C. \(40 \text{ m/s}^2\)
• D. \(35 \text{ m/s}^2\)

Hint:

Remember that velocity is a vector; since the direction changes, you must add the magnitudes to find the total change in velocity.

Answer 1

The Correct Option is A

Step 1: Velocity before impact
$v_1 = \sqrt{2gh_1} = \sqrt{2 \cdot 10 \cdot 20} = 20 \text{ m/s}$ (downward).

Step 2: Velocity after impact
$v_2 = \sqrt{2gh_2} = \sqrt{2 \cdot 10 \cdot 5} = 10 \text{ m/s}$ (upward).

Step 3: Average Acceleration
$a = \frac{\Delta v}{\Delta t} = \frac{v_2 - (-v_1)}{t} = \frac{10 + 20}{1} = 30 \text{ m/s}^2$.
Final Answer: (A)