S=100m... u=0m/s... g=10m/s^2 By newton's 3rd eqn of motion. v^2=(0)^2+(2.10.100)... v^2= 2000 so v=20(rt5)m/s..

Introduction:
When a stone is dropped from a cliff, it falls freely under the influence of gravity. The stone's speed gradually increases as it falls due to the acceleration caused by gravity. After falling a certain distance, in this case, 100 meters, we can calculate the stone's speed using the equations of motion.

Calculating the Speed:
To calculate the speed of the stone after falling 100 meters, we can use the equation:

v^2 = u^2 + 2as

Where:
- v is the final velocity (speed) of the stone
- u is the initial velocity (which is zero since the stone is dropped)
- a is the acceleration due to gravity (approximately 9.8 m/s^2)
- s is the distance fallen (100 meters)

Substituting the Values:
Let's substitute the given values into the equation and solve for v:

v^2 = 0^2 + 2 * 9.8 * 100

Simplifying the equation:

v^2 = 1960

Taking the square root of both sides to solve for v:

v = √1960

v ≈ 44.27 m/s

Conclusion:
After falling 100 meters, the stone's speed would be approximately 44.27 m/s. It is important to note that this calculation assumes no air resistance or other external factors affecting the stone's motion.