A ball is dropped from a height h above ground. Neglect the air resistance, its velocity (v) varies with its height (y) above the ground as :-

Options

A. `sqrt(2g(h-y))`
B. `sqrt(2gh)`
C. `sqrt(2gy)`
D. `sqrt(2g(h+y))`

Answer

To solve the problem of how the velocity (v) of a ball varies with its height (y) above the ground when dropped from a height (h), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The ball is dropped from a height \( h \) above the ground.
- The initial velocity \( u \) of the ball when it is dropped is \( 0 \) m/s.

2. **Define the Variables:**
- Let \( y \) be the height of the ball above the ground at any time.
- The distance fallen by the ball from the height \( h \) to height \( y \) is \( h - y \).

3. **Use the Kinematic Equation:**
- We will use the kinematic equation that relates velocity, initial velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
- Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance fallen.

4. **Substitute Known Values:**
- Since the ball is in free fall, the acceleration \( a \) is equal to the acceleration due to gravity \( g \).
- The initial velocity \( u = 0 \).
- The distance \( s \) is \( h - y \).
- Substituting these values into the equation gives:
\[ v^2 = 0 + 2g(h - y) \]
- This simplifies to:
\[ v^2 = 2g(h - y) \]

5. **Solve for Velocity \( v \):**
- Taking the square root of both sides, we get:
\[ v = \sqrt{2g(h - y)} \]

6. **Final Expression:**
- Thus, the velocity \( v \) of the ball as a function of its height \( y \) above the ground is:
\[ v = \sqrt{2g(h - y)} \]