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?

hfinal=hinitial+Vi*t-4.9t^2

you have hfinal=0, t, Vi=15sin30, solve for hinitial

Where does the 4.9 come from though?

1/2 g= 4.9

Oh, is it .5 multiplied by the acceleration of gravity?

yes.

Alright. Thank you so much for your help!

To find the height of the cliff, we can use the equations of motion for projectile motion.

First, we need to break down the initial velocity into its horizontal and vertical components.

Given:
Initial velocity (v) = 15 m/s
Angle of launch (θ) = 30 degrees

The horizontal component of the initial velocity can be calculated as:
Vx = v * cos(θ)

The vertical component of the initial velocity can be calculated as:
Vy = v * sin(θ)

Now we can use the equation of motion to calculate the time it takes for the rock to reach the ground. The equation for vertical displacement is:
y = Vy * t + (1/2) * g * t^2

where:
y = vertical displacement (height of the cliff)
t = time of flight (6.30 seconds)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Rearranging the equation, we get:
y = (Vy * t) + (1/2) * g * t^2

Substituting the value of Vy and g, we have:
y = (v * sin(θ) * t) + (1/2) * (9.8) * t^2

Now we can substitute the known values into the equation:
y = (15 * sin(30) * 6.30) + (0.5 * 9.8 * (6.30)^2)

Simplifying and calculating the equation, we get:
y = 94.5 + 294.14

Therefore, the height of the cliff is:
y = 388.64 meters.

Hence, the cliff is approximately 388.64 meters high.