A rock is thrown upwards from the top of a 100m cliff, on the way down the rock misses the edge of the cliff and continues falling to the ground. What is the distance that the rock travelled?

h = v²/2g

Distance= 2h+100(m)
The speed is necessary

To find the distance the rock traveled, we need to break down the motion into two parts: the upward motion and the downward motion.

First, let's find the distance traveled during the upward motion. We know the initial velocity of the rock is zero when it is thrown upwards, and it eventually reaches its maximum height when its vertical velocity becomes zero again. Since we only need to consider the distance traveled, we don't need to consider any horizontal motion.

We can use the following kinematic equation to find the distance traveled during the upward motion:
s = ut + (1/2)at^2

Where:
- s is the distance traveled
- u is the initial velocity (which is 0 in this case)
- t is the time taken during the upward motion
- a is the acceleration due to gravity (approximately -9.8 m/s^2)

We don't have the value of time for the upward motion, so we need to find that first. We can use the following equation to find the time taken to reach the maximum height:
v = u + at

Where:
- v is the final velocity (which is 0)
- u is the initial velocity (which is 0)
- a is the acceleration due to gravity (approximately -9.8 m/s^2)
- t is the time taken

Rearranging the equation, we have:
t = -u/a

Since the initial velocity is 0, the time taken for the upward motion is 0.

Now, let's find the distance traveled during the downward motion. Since the rock is falling under the influence of gravity, the distance traveled can be calculated using another kinematic equation:
s = ut + (1/2)at^2

Where:
- s is the distance traveled
- u is the initial velocity (which is 0 for the downward motion because it starts from rest)
- t is the time taken for the downward motion
- a is the acceleration due to gravity (approximately -9.8 m/s^2)

Again, we don't have the value of time, but we can find it by using the following equation:
-h = ut + (1/2)at^2

Where:
- h is the height of the cliff (100 m)
- u is the initial velocity (which is 0)
- t is the time taken for the rock to fall from the top of the cliff
- a is the acceleration due to gravity (approximately -9.8 m/s^2)

Rearranging the equation, we get:
t = sqrt((2h)/g)

Now we have the time for the downward motion. Let's substitute the values into the distance equation:
s = 0t + (1/2)(-9.8)t^2

Simplifying the equation and plugging in the values, we get:
s = (1/2)(-9.8)(sqrt((2h)/g))^2

Substituting h = 100m and g = 9.8 m/s^2, we can calculate the distance traveled during the downward motion.