a rock is dropped from a height of 15m above the ground. when it hits the ground, it leaves a hole 12cm deep. what is the speed of the rock when it hits the ground? what is the deceleration of the rock when it hits the ground?
See your 4:26pm post for solution.
some massive butt mud
To find the speed of the rock when it hits the ground, we can use the principle of conservation of energy. Initially, the only form of energy the rock possesses is potential energy due to its height above the ground. As it falls, this potential energy is converted into kinetic energy.
The potential energy (PE) of an object at a certain height is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
In this case, the height (h) is 15 meters. However, we don't have the mass of the rock, so we need to find another way to determine its speed.
We can use the information about the depth of the hole created by the rock to find the speed. When the rock hits the ground, all of its initial potential energy (PE) is converted to kinetic energy (KE). Therefore, we can equate the potential energy at the height with the kinetic energy at the bottom of the drop.
PE = KE
mgh = (1/2)mv^2
Here, v represents the final velocity/speed of the rock. We can solve for v by rearranging the equation:
v^2 = 2gh
v = ā(2gh)
Now, let's substitute the values:
g ā 9.8 m/s^2
h = 15 meters
v = ā(2 * 9.8 * 15)
After calculating this expression, we find that the speed of the rock when it hits the ground is approximately 17.1 m/s.
Moving on to the deceleration, we need more information about the time it takes for the rock to come to rest to calculate it. Could you please provide the stopping time of the rock?