a 20 kg rock is on the edge of a 100 m cliff . the rock falls from the cliff , what is its kinetic energy just before it hits the ground ?

can you give me the answer

Well, well, well, looks like the rock is ready to rock and roll, quite literally. Alright, let's do some calculations. To find the kinetic energy, we need to know the speed of that rock just before it goes splat on the ground.

Now, since our rock is falling, we can use the height of the cliff, which is 100 m, to find the speed just before it hits the ground. According to my calculations, it will be approximately 44.3 m/s.

Now, the formula to calculate kinetic energy is KE = 1/2 * mass * velocity^2. Plugging in our numbers, we get KE = 1/2 * 20 kg * (44.3 m/s)^2. Crunching the numbers, we get a whopping 19,558 Joules of kinetic energy.

So, the rock's kinetic energy just before it hits the ground is approximately 19,558 Joules. That's some serious rock and roll energy, my friend!

To calculate the kinetic energy of the rock just before hitting the ground, we need to know its final velocity. We can calculate this using the conservation of energy principle, considering the potential energy of the rock at the top of the cliff and its kinetic energy just before hitting the ground.

The potential energy (PE) of the rock at the top of the cliff is given by the formula:

PE = m * g * h

Where:
m = mass of the rock = 20 kg (given)
g = acceleration due to gravity = 9.8 m/s^2 (approx.)
h = height of the cliff = 100 m (given)

Substituting the given values, we get:

PE = 20 kg * 9.8 m/s^2 * 100 m
= 19600 J

According to the law of conservation of energy, this potential energy is converted into kinetic energy (KE) just before the rock hits the ground. Therefore, the kinetic energy of the rock is also 19600 J.

So, the kinetic energy of the rock just before hitting the ground is 19600 Joules.

Is basically 9.8 times 100 times 20 thats ur answer fool

M*g*H

H = 100 m
M = 20 kg
g = 9.8 m/s^2

The answer will be in Joules.