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. \(39.6\,\text{m/s}\)

Hint:

When an object is 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

Step 1: Understanding the Question:
We need to find the final impact velocity of an object falling from rest under the influence of gravity from a known height.
Step 2: Key Formula or Approach:
Using the third equation of motion, \(v^2 = u^2 + 2gh\). Since the ball is "dropped", the initial velocity \(u = 0\). This simplifies to:
\[ v = \sqrt{2gh} \]
where \(g\) is the acceleration due to gravity (taken as \(9.8\,\text{m/s}^2\)) and \(h\) is the vertical distance.
Step 3: Detailed Explanation:
Substitute the given height \(h = 20\,\text{m}\) and \(g = 9.8\,\text{m/s}^2\) into the equation:
\[ v = \sqrt{2 \times 9.8 \times 20} \]
\[ v = \sqrt{19.6 \times 20} \]
\[ v = \sqrt{392} \]
Calculating the square root:
\[ v \approx 19.8\,\text{m/s} \]
Step 4: Final Answer:
The velocity of the ball just before impact is approximately \(19.8\,\text{m/s}\).