A rock is thrown vertically upward from the edge of a cliff. The rock reaches a maximum height of 20.6 m above the top of the cliff before falling to the base of the cliff, landing 6.54 s after it was thrown. How high is the cliff?

V^2 = Vo^2 + 2g*d.

Vo^2 = V^2--2g*d
Vo^2 = 0 - (-19.6)*20.6 = 403.76
Vo = 20.1 m/s. = Initial velocity.

V = Vo + g*t.
Tr = (V-Vo)/g = (0-20.1)/-9.8 = 2.05 s.
= Rise time.

Tr + Tf = 6.54 s.
2.05 + Tf = 6.54
Tf = 6.54 - 2.05 = 4.49 s. = Fall time
from 20.6 m above the cliff to Gnd.

Vo*t + 0.5g*t^2 = 20.6 + h.
0 + 4.9t^2 = 20.6 + h
h = 4.9*(4.49)^2 - 20.6 = 78.2 m.

To determine the height of the cliff, we can use the equations of motion to analyze the path of the rock. Let's break down the problem into different stages.

Stage 1: The rock is thrown vertically upward.
During the first part of the motion, the rock goes up, reaches its maximum height, and begins to fall back down.

Given:
Maximum height (H) = 20.6 m
Time taken to reach maximum height (t) = ? (unknown)
Acceleration due to gravity (g) = -9.8 m/s² (negative sign because it acts downward)

We know that the velocity of the rock when it reaches its maximum height is zero since it momentarily comes to rest before falling back down. We can use the equation of motion:

v = u + gt

Here, u represents the initial velocity, v represents the final velocity, g is the acceleration due to gravity, and t is the time taken.

Since v = 0 when the rock reaches its maximum height, the equation becomes:

0 = u + (-9.8)t

Rearranging the equation, we find:

-9.8t = u

Since the rock is thrown vertically upward, the initial velocity is u = u₀ (initial vertical velocity).

Stage 2: The rock falls back to the base of the cliff.
During this stage, the rock falls downward until it reaches the base of the cliff.

Given:
Time taken to reach the base (t) = 6.54 s
Acceleration due to gravity (g) = -9.8 m/s²

To find the height of the cliff, we need to determine the time taken for the rock to reach its maximum height. Let's calculate that first.

From the first stage, we know that -9.8t = u, so t = u/-9.8.

Using this time value, we can set up an equation for the total time of flight (t_total):

t_total = t + t

Substituting the values:

t_total = (u/-9.8) + 6.54

Since the rock reaches its maximum height when v = 0, we can use the equation:

v = u + gt

0 = u + (-9.8)t
9.8t = u

Substituting this value of u in the equation for t_total:

t_total = (9.8t / -9.8) + 6.54

Simplifying:

t_total = t + 6.54

Since the total time of flight is the sum of the time taken to reach the maximum height and the time taken to fall back down, we now have an equation to solve for t_total.

Using the quadratic formula:

t_total = t + 6.54
0 = -t_total^2 + 6.54t_total - 6.54t

Solving this quadratic equation will give you the value of t_total.

Once you have the value of the total time of flight (t_total), you can use it to find the height of the cliff by using the equation:

H = ut_total + (1/2)gt_total^2

Now, substitute the known values (H = 20.6m, g = -9.8m/s², and t_total):

20.6 = u(t_total) + (1/2)(-9.8)(t_total)^2

Finally, solve this equation to find the initial vertical velocity (u₀):

u₀ = (20.6 - (1/2)(-9.8)(t_total)^2) / t_total

Then, substitute the value of u₀ into the equation for the height of the cliff:

H_cliff = u₀t_total + (1/2)gt_total^2

Calculate and simplify to find the height of the cliff.