Question

A ball is dropped from a height of \(20\,\text{m}\). What is its velocity just before hitting the ground?

• A. \(14\,\text{m/s}\)
• B. \(19.8\,\text{m/s}\)
• C. \(9.8\,\text{m/s}\)
• D. \(39.6\,\text{m/s}\) \bigskip

Hint:

When an object is \textbf{dropped}, the initial velocity is \(0\). In such cases, the velocity before hitting the ground can be directly found using \(v=\sqrt{2gh}\).

Answer 1

The Correct Option is B

Concept: When an object is dropped from a height, it falls freely under the influence of gravity. The velocity of the object just before hitting the ground can be determined using the kinematic equation: :contentReference[oaicite:0]{index=0} where \begin{itemize} \item \(v\) = final velocity just before impact \item \(g\) = acceleration due to gravity \((9.8\,\text{m/s}^2)\) \item \(h\) = height from which the object is dropped \end{itemize} Step 1: {\color{red}Substitute the given values into the formula.} \[ v = \sqrt{2 \times 9.8 \times 20} \] Step 2: {\color{red}Perform the calculation.} \[ v = \sqrt{392} \] \[ v \approx 19.8\,\text{m/s} \] Thus, the velocity of the ball just before hitting the ground is: \[ v = 19.8\,\text{m/s} \] \bigskip