A rock is thrown from on a cliff with the velocity of 15 m/s at an angle of 30 degrees. it takes the rock 6.30 seconds to hit the ground. how high was the cliff?

a = V/T = 15 / 6.3 = 2.38 m/s^2.

d = Vo*t + 0.5at^2,
= 15*6.3 + 0.5*2.38(6.3)^2,
= 94.5 + 47.2 = 141.7 m @ 30 Deg.

h = 141.7sin30 = 70.9 m.

To determine the height of the cliff, we can use the equations of motion and consider the vertical motion of the rock. We need to break down the initial velocity of the rock into its vertical and horizontal components.

Given:
Initial velocity, v = 15 m/s
Launch angle, θ = 30 degrees
Time of flight, t = 6.30 s

Step 1: Break down the initial velocity into vertical and horizontal components.
The vertical component of velocity (Vy) can be calculated using the formula:
Vy = v * sin(θ)

Substituting the given values:
Vy = 15 * sin(30)
Vy ≈ 7.5 m/s

Step 2: Determine the maximum height reached by the rock during its flight. At the maximum height, the vertical component of the rock's velocity becomes zero (Vy = 0). The time taken to reach the maximum height is half of the total time of flight (t/2).

Using the equation:
Vy = u + at

Where u is the initial vertical velocity, a is acceleration (which is -9.8 m/s^2 for objects falling downward), and t is the time taken.

Substituting the given values:
0 = 7.5 - 9.8 * (t/2)

Solving for t:
t/2 = 7.5 / 9.8
t ≈ 1.53 s

Step 3: Calculate the maximum height (h) using the equation of motion:
h = u * t + (1/2) * a * t^2

Substituting the values:
h = 7.5 * 1.53 + (1/2) * (-9.8) * (1.53)^2
h ≈ 5.74 meters

Therefore, the height of the cliff is approximately 5.74 meters.