Question

A body is falling freely from the top of a tower of height 125 m. The distance covered by the body during the last second of its motion is \(x\%\) of the height of the tower. Then \(x\) is (Acceleration due to gravity = 10 m/s\(^2\))

• A. \(9\%\)
• B. \(36\%\)
• C. \(25\%\)
• D. \(49\%\)

Hint:

Remember that the distance fallen in the last second can be found using the difference in distances from consecutive seconds.

Answer 1

The Correct Option is B

The total height \(h = 125\) m and acceleration due to gravity \(g = 10\) m/s\(^2\). Using the formula for the distance fallen in the last second \(h = \frac{1}{2}g(t-1)^2 - \frac{1}{2}gt^2\): \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 125}{10}} = 5 \text{ seconds} \] Distance fallen in the last second: \[ h = \frac{1}{2}g(5-1)^2 - \frac{1}{2}g(5)^2 = 45 \text{ m} \] Thus, \(x = \frac{45}{125} \times 100 = 36\%\).