A stone is aimed at a cliff of height h with an initial speed of v = 54.0 m/s directed 55.0° above the horizontal, as shown in the Figure below by the arrow. The stone strikes at A, 7.31 s after launching. What is the height of the cliff?

What is the maximum height H reached above the ground?

To calculate the height of the cliff, we can use the equation of motion in the vertical direction:

h = v₀y * t + (1/2) * g * t²

where:
h = height of the cliff
v₀y = initial vertical velocity
t = time
g = acceleration due to gravity

First, we need to find the initial vertical velocity, v₀y. We can use the given initial speed, v, and the angle of projection, θ:

v₀y = v * sin(θ)

Therefore, v₀y = 54.0 m/s * sin(55.0°) = 44.05 m/s

Now we can substitute the values into the equation to find the height of the cliff:

h = (44.05 m/s) * (7.31 s) + (1/2) * (-9.8 m/s²) * (7.31 s)²

h ≈ 44.05 m/s * 7.31 s - 0.5 * 9.8 m/s² * (7.31 s)²
h ≈ 321.2565 m - 259.08653 m
h ≈ 62.17 m

Therefore, the height of the cliff is approximately 62.17 meters.

To find the maximum height reached above the ground, we can use the equation:

H = v₀y² / (2 * g)

Substituting the values:

H = (44.05 m/s)² / (2 * 9.8 m/s²)
H ≈ 968.9025 m²/s² / 19.6 m/s²
H ≈ 49.45 m

Therefore, the maximum height reached above the ground is approximately 49.45 meters.

To find the height of the cliff, we can use the kinematic equation for vertical motion.

The equation is: h = v0y * t + 0.5 * g * t^2

Where:
- h is the vertical displacement or height of the cliff
- v0y is the vertical component of the initial velocity
- t is the time of flight
- g is the acceleration due to gravity

We are given:
- v = 54.0 m/s (initial speed)
- θ = 55.0° (angle above the horizontal)
- t = 7.31 s (time of flight)
- g = 9.8 m/s^2 (acceleration due to gravity)

First, we need to find the vertical component of the initial velocity (v0y). We can use trigonometry to find this:

v0y = v * sin(θ)
v0y = 54.0 * sin(55.0°)
v0y ≈ 44.12 m/s

Now, we can substitute the values into the equation for h:

h = v0y * t + 0.5 * g * t^2
h = 44.12 * 7.31 + 0.5 * 9.8 * (7.31^2)
h ≈ 157.29 m

Therefore, the height of the cliff is approximately 157.29 meters.

To find the maximum height, we can use the fact that at the highest point, the vertical velocity is zero (v = 0).

We can use the following equation to find the time (tmax) it takes for the stone to reach its maximum height:

0 = v0y - g * tmax

Solving for tmax, we have:

tmax = v0y / g
tmax = 44.12 / 9.8
tmax ≈ 4.50 s

Now, we can substitute the value of tmax into the equation for vertical displacement:

H = v0y * tmax - 0.5 * g * (tmax^2)
H = 44.12 * 4.50 - 0.5 * 9.8 * (4.50^2)
H ≈ 99.68 m

Therefore, the maximum height reached above the ground is approximately 99.68 meters.