To find the height of the building, we can use the equations of motion in the vertical direction.
First, let's find the time it takes for the rock to hit the ground. We can use the equation:
y = y0 + V0y*t - (1/2)*g*t^2
where:
- y is the vertical displacement (height) of the rock
- y0 is the initial vertical position (height) of the rock (which is the height of the building)
- V0y is the initial vertical component of the velocity
- g is the acceleration due to gravity
- t is the time taken
Since the initial vertical velocity is given by V0y = V0*sin(θ), where V0 is the initial velocity and θ is the angle above the horizontal, we can rewrite the equation as:
y = y0 + V0*sin(θ)*t - (1/2)*g*t^2 --------(1)
The horizontal distance traveled by the rock is given by:
x = V0x*t
where V0x is the initial horizontal component of the velocity. In this case, V0x = V0*cos(θ).
We are given that x = 19.5 m, V0 = 11.6 m/s, θ = 52°, and g = 9.8 m/s^2. Substituting these values into the equation, we get:
19.5 = (11.6*cos(52°))*t --------(2)
Now, we can solve equations (1) and (2) simultaneously to find the height of the building (y0) and the time (t).
From equation (2):
t = 19.5 / (11.6*cos(52°))
Substituting this value of t into equation (1), we can solve for y0.
y = y0 + V0*sin(θ)*t - (1/2)*g*t^2
y0 = y - V0*sin(θ)*t + (1/2)*g*t^2
Substituting the known values, we have:
y0 = y - 11.6*sin(52°)*(19.5 / (11.6*cos(52°))) + (1/2)*9.8*(19.5 / (11.6*cos(52°)))^2
Evaluating this expression will give us the height of the building.