On Earth a rock falls an unknown vertical distance from a resting position and lands in a lake. If it takes the rock 2.5 seconds to fall, how high is the cliff that the rock fell from?

the formula for vertical distance is,

d = (v0)*t-(1/2)g*t^2
where v0=initial velocity, t=time, and g=acceleration due to gravity, which is approximately 9.8 m/(s^2)
*now since this is freefall, v0 = 0 and the formula reduces to:
d = -(1/2)g*t^2
d = -(1/2)*9.8*(2.5^2)

then solve for d. units in m. (get its absolute value since it's distance)

THANK YOU =0]

On Earth a rock falls an unknown vertical distance from a resting position and lands in a lake. If it takes the rock 2.5 seconds to fall, how high is the cliff that the rock fell from?

To determine the height of the cliff, we can use the equation of motion for free-falling objects. This equation is given as:

h = (1/2) * g * t^2

Where:
- h is the height of the cliff
- g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
- t is the time it takes for the rock to fall (given as 2.5 seconds)

Plugging the values into the equation, we can find the height of the cliff:

h = (1/2) * 9.8 * (2.5^2)
= 0.5 * 9.8 * 6.25
= 30.625 meters

Therefore, the height of the cliff from which the rock fell is approximately 30.625 meters.