Question:medium

A stone is dropped from a height of \(20\,\text{m}\); calculate its velocity just before it hits the ground \((g = 10\,\text{m/s}^2)\).

Show Hint

When an object is dropped freely, the initial velocity \(u = 0\). Use the equation \(v^2 = 2gh\) to quickly find the final velocity.
Updated On: Apr 18, 2026
  • \(10\,\text{m/s}\)
  • \(15\,\text{m/s}\)
  • \(20\,\text{m/s}\)
  • \(25\,\text{m/s}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
We need to find the final velocity (\(v\)) of a stone that is dropped from a certain height (\(h\)), considering the acceleration due to gravity (\(g\)).
Step 2: Key Formula or Approach:
For an object in free fall under constant acceleration, we can use the third equation of motion. This equation relates final velocity (\(v\)), initial velocity (\(u\)), acceleration (\(g\)), and displacement (\(h\)):
\[ v^2 = u^2 + 2gh \] Step 3: Detailed Explanation:
(i) Identify the given quantities:
- The stone is "dropped," which implies its initial velocity is zero: \(u = 0 \, \text{m/s}\).
- The height from which it is dropped: \(h = 20 \, \text{m}\).
- The acceleration due to gravity: \(g = 10 \, \text{m/s}^2\).
(ii) Substitute the values into the kinematic equation:
\[ v^2 = (0)^2 + 2(10)(20) \] \[ v^2 = 0 + 400 \] \[ v^2 = 400 \] (iii) Solve for the final velocity, v:
\[ v = \sqrt{400} \] \[ v = 20 \, \text{m/s} \] Step 4: Final Answer:
The velocity of the stone just before it hits the ground is \(20\,\text{m/s}\).
Was this answer helpful?
2