Question

A stone is dropped from the top of a tower. The height through which it falls in the first 3 seconds of its motion equals the height through which it falls in the last second of its motion. To reach the ground, the stone takes time equal to (g = 10 m/s\(^2\)):

• A. 4 s
• B. 5 s
• C. 6 s
• D. 7 s

Hint:

The distance fallen in the last second of motion is equal to the difference in the heights fallen at time \( t \) and \( t-1 \).

Answer 1

The Correct Option is C

Let the total time taken to reach the ground be \( t \). The height fallen in the first 3 seconds is given by: \[ h_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} \times 10 \times 3^2 = 45 \, \text{m} \] The distance fallen in the last second is given by the difference of the distances fallen in \( t \) seconds and \( t-1 \) seconds: \[ h_2 = \frac{1}{2} g t^2 - \frac{1}{2} g (t-1)^2 = 45 \, \text{m} \] Solving this equation gives \( t = 6 \, \text{s} \).