OnlineCBT
Physics · MOTION IN A STRAIGHT LINE
हिन्दीA ball is dropped from a height of \( 90\text{ m} \) on a floor. Taking \( g = 9.8\text{ m/s}^2 \), if the ball loses one-tenth of its speed at each collision with the floor, what is the speed of the ball just before the first collision and the height to which it rebounds after the first bounce?
Correct answer · Option 1
\( 42\text{ m/s} \text{ and } 72.9\text{ m} \)
Explanation
Speed before bounce \( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 90} = 42\text{ m/s} \). Speed after losing 1/10th speed \( v' = 0.9 \times 42 = 37.8\text{ m/s} \). Rebound height \( h' = \frac{(v')^2}{2g} = \frac{(37.8)^2}{2 \times 9.8} = 72.9\text{ m} \).
