Question

A ball is dropped from a height \(18\,\text{m}\) above the ground. At any moment, if magnitude of acceleration and velocity are same, find its height above ground at that instant.

• A. \(15\ \text{m}\)
• B. \(5\ \text{m}\)
• C. \(13\ \text{m}\)
• D. \(2\ \text{m}\)

Answer 1

The Correct Option is C

Concept: For free fall, \[ v = gt \] Acceleration remains constant: \[ a = g \] \includegraphics[width=0.5\linewidth]{9p ans.png} Step 1: Condition given Magnitude of velocity equals acceleration. \[ v = a \] \[ gt = g \] \[ t = 1\ \text{s} \] Step 2: Distance fallen \[ s = \frac12 gt^2 \] \[ s = \frac12 (10)(1)^2 \] \[ s = 5\ \text{m} \] Step 3: Height above ground Initial height \[ H = 18\ \text{m} \] \[ h = 18 - 5 \] \[ h = 13\ \text{m} \]