Arock is dropped from a cliff and hits the ground 6.0 seconds later. how high is the cliff?

The height of the cliff is found to be 176.4 m

Explanation:

Since, the body is being dropped from a certain height and reaches the ground in some time. Thus, we can apply the equations of motion (modified for vertical motion) in this case, due to constant accelerated motion. We have the following data:

Acceleration due to gravity = g = 9.8 m/s²

Time to reach ground = t = 6 sec

Initial Velocity of the Rock = Vi = 0 m/s (Because, the rock will be at rest, initially)

Height of cliff = H = ?

Now, applying second equation of motion (modified for vertical motion), to the rock, between the top of the cliff and ground, we get:

H = Vi t + (1/2)gt²

Using values:

H = (0 m/s)(6 sec) + (1/2)(9.8 m/s²)(6 sec)²

H = 176.4 m

i think the answer is a and b

i wish you luck

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2