1. Given data

• Spring constant: $k = 100\,\text{N/m}$
• Mass of ball: $m = 100\,\text{g} = 0.1\,\text{kg}$
• Compression of spring: $x = 0.05\,\text{m}$
• Height from ground: $h = 2\,\text{m}$
• Acceleration due to gravity: $g = 10\,\text{m/s}^2$

The ball leaves the gun horizontally.

2. Find the speed of the ball as it leaves the spring gun

Using conservation of energy:

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

Substitute the values:

$\frac{1}{2}(100)(0.05)^2 = \frac{1}{2}(0.1)v^2$

$50 \times 0.0025 = 0.05v^2$

$0.125 = 0.05v^2$

$v^2 = \frac{0.125}{0.05} = 2.5$

$v = \sqrt{2.5}\,\text{m/s}$

3. Find the time of fall from height $2\,\text{m}$

Since the ball is projected horizontally, initial vertical velocity is zero.

$h = \frac{1}{2}gt^2$

$2 = \frac{1}{2}(10)t^2$

$2 = 5t^2$

$t^2 = 0.4$

$t = \sqrt{0.4}\,\text{s}$

4. Find horizontal distance $d$

$d = vt$

$d = \sqrt{2.5} \cdot \sqrt{0.4}$

$d = \sqrt{2.5 \times 0.4} = \sqrt{1} = 1\,\text{m}$

5. Final answer

$\boxed{d = 1\,\text{m}}$

This matches the stored correct answer.