A 0.67 kg rock is projected from the edge of
the top of a building with an initial velocity of
11.6 m/s at an angle 52
◦
above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 19.5 m from the base
of the building.
How tall is the building? Assume the
ground is level and that the side of the building is vertical. The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m
Well, let me calculate it for you, but first I must ask: Why did the rock go on a trip like that? Did it have a falling out with the building and decide to go on vacation?
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.
To find the height of the building, we need to break the motion of the rock into its horizontal and vertical components.
First, let's calculate the time it takes for the rock to hit the ground. We know that the horizontal distance traveled by the rock is 19.5 m, and the initial horizontal velocity is 11.6 m/s. The time can be calculated using the equation:
distance = velocity * time
19.5 m = 11.6 m/s * time
Solving this equation, we find that the time it takes for the rock to travel 19.5 m horizontally is approximately 1.68 seconds.
Now, let's calculate the vertical distance traveled by the rock. We can use the equation for vertical displacement:
vertical distance = initial vertical velocity * time - 1/2 * acceleration due to gravity * time^2
The initial vertical velocity can be calculated using the initial velocity and the angle of projection. We can use the sine function to find it:
initial vertical velocity = initial velocity * sin(angle)
initial vertical velocity = 11.6 m/s * sin(52 degrees)
Using a scientific calculator, we find that the initial vertical velocity is approximately 9.07 m/s.
Substituting the values into the equation for vertical distance, we get:
vertical distance = 9.07 m/s * 1.68 s - 1/2 * 9.8 m/s^2 * (1.68 s)^2
Simplifying this equation, we find that the vertical distance traveled by the rock is approximately 12.16 m.
Finally, to find the height of the building, we add the vertical distance traveled by the rock to the height from which it was projected:
height = vertical distance + initial height
The initial height is the height of the building. Therefore,
height = 12.16 m + 0 m
So, the height of the building is approximately 12.16 m.
Upward motion of projectile to the top point
The height of this point is
hₒ= vₒ²•sin²α/2g = 11.6²•sin²52/2•9.8 = 4.26 m.
horizontal displacement is
L1 = vₒ²•sin2 α/2g=6.6 m.
Downward motion with horizontal velocity
v(x) = vₒ•cosα =11.6•cos52=7.14 m/s.
The horizontal displacement at this motion is
L2 = L-L1 =19.5-6.66=12.84 m.
The time of the motion is
t=L2/v(x) = 12.84/7.14 = 1.79 s.
The vertical displacement is
H=gt²/2 =9.8•1.79²/2 = 15.74 m.
The height of the building is
h=H-hₒ17.74-4.26 = 11.48 m.