Arock is dropped from the edge of a cliff into a pool of water. assume free-fall acceleration is 10 m/s per second, and air resistance can be neglected. if the rock falls for 2.0 seconds before it hits the water, the height of the cliff above the pool is

20 m

Explanation:

From the equation of motion,

S = ut+1/2gt² Equation 1

Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.

Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative

Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)

Substituting into equation 1

S = 0(2) + 1/2(10)(2)²

S = 5(4)

S = 20 m

Hence the height of the the cliff above the pool is 20 m

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr