A 0.65 kg rock is projected from the edge of
the top of a building with an initial velocity of
10.5 m/s at an angle 40◦
above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 19.6 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
I could not care less how massive the rock is.
Hi = height of building
Vi = initial v up = 10.5 sin 40 = 6.75 m/s
u = constant horizontal speed = 10.5 cos 40 = 8.04 m/s
x = 19.6 = u t = 8.04 t
so t = time aloft = 19.6 / 8.04 = 2.44 seconds
h = Hi + Vi t - 4.9 t^2
0 = Hi + 6.75 * 2.44 - 4.9 * 2.44^2
solve for Hi
Well, it seems like the rock has quite a "fall-ing" love for gravity! Let's help it find its height.
First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component is given by the equation: Vx = V * cos(theta), where V is the initial velocity and theta is the angle above the horizontal. Plugging in the values, Vx = 10.5 m/s * cos(40°).
The vertical component is given by the equation: Vy = V * sin(theta). Plugging in the values, Vy = 10.5 m/s * sin(40°).
Now, let's find how long the rock takes to reach the ground. We can use the equation: t = 2 * Vy / g, where g is the acceleration due to gravity. Plugging in the values, t = 2 * (10.5 m/s * sin(40°)) / 9.8 m/s^2.
Next, let's find the height of the building. We can use the equation: h = Vy * t + 0.5 * g * t^2, where Vy is the vertical component of velocity and t is the time. Plugging in the values, h = (10.5 m/s * sin(40°)) * t + 0.5 * 9.8 m/s^2 * t^2.
Finally, we know that the horizontal distance traveled by the rock is given by the equation: d = Vx * t. We can plug in the values and solve for t: 19.6 m = (10.5 m/s * cos(40°)) * t.
Now that we have the value of t, we can substitute it back into the equation for h to find the height of the building. Solving the equation, we will get the answer in units of meters.
Calculations aside, it seems like this rock had a real "gravity" adventure!
To find the height of the building, we can use the equation of motion for projectile motion in the vertical direction. The equation is as follows:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
y = vertical displacement (height of the building)
y0 = initial vertical position
v0y = vertical component of the initial velocity
g = acceleration due to gravity
t = time of flight
We need to find the time of flight first. The time of flight can be calculated using the horizontal distance traveled by the rock:
x = v0x * t
Where:
x = horizontal distance traveled by the rock
v0x = horizontal component of the initial velocity
t = time of flight
Given:
v0 = 10.5 m/s
θ = 40°
x = 19.6 m
g = 9.8 m/s^2
We can calculate the horizontal and vertical components of the initial velocity as follows:
v0x = v0 * cos(θ)
v0y = v0 * sin(θ)
Now, let's calculate v0x and v0y:
v0x = 10.5 m/s * cos(40°)
v0x ≈ 10.5 m/s * 0.766
v0x ≈ 8.03 m/s
v0y = 10.5 m/s * sin(40°)
v0y ≈ 10.5 m/s * 0.643
v0y ≈ 6.76 m/s
Next, let's find the time of flight:
x = v0x * t
19.6 m = 8.03 m/s * t
t = 19.6 m / 8.03 m/s
t ≈ 2.44 s
Now that we have the time of flight, we can use it to find the height of the building:
y = y0 + v0y * t - (1/2) * g * t^2
Since the rock was projected from the edge of the top of the building, the initial vertical position (y0) is equal to the height of the building.
y = h, where h is the height of the building.
h ≈ y + v0y * t - (1/2) * g * t^2
h ≈ 0 + 6.76 m/s * 2.44 s - (1/2) * 9.8 m/s^2 * (2.44 s)^2
Now, let's calculate h:
h ≈ 16.47 m
Therefore, the height of the building is approximately 16.47 meters.
To find the height of the building, you need to determine the time it takes for the rock to reach the ground. Once you have the time, you can use the equation for height under constant acceleration to calculate the height.
First, split the initial velocity of the rock into its horizontal and vertical components. The horizontal component is given by Vx = V_initial * cos(theta), where theta is the angle above the horizontal (40 degrees in this case), and V_initial is the initial velocity (10.5 m/s). The vertical component is given by Vy = V_initial * sin(theta).
Next, determine the time it takes for the rock to reach the ground. Since there is no vertical acceleration in the horizontal direction, the time of flight can be found using the equation:
time = horizontal distance / horizontal velocity.
In this case, the horizontal distance is 19.6 m and the horizontal velocity is Vx.
Now, you can use the equation for height under constant acceleration to find the height of the building. The equation is:
h = Vy * t + (1/2) * g * t^2.
Substituting the values you have, where h is the height of the building, Vy is the vertical component of the initial velocity, t is the time of flight, and g is the acceleration due to gravity (9.8 m/s^2), you can solve for h.
Let's calculate it step by step:
1. Calculate horizontal component velocity:
Vx = V_initial * cos(theta)
Vx = 10.5 m/s * cos(40 degrees)
2. Calculate time of flight:
time = horizontal distance / horizontal velocity
time = 19.6 m / Vx
3. Calculate the height of the building:
h = Vy * t + (1/2) * g * t^2
h = V_initial * sin(theta) * time + (1/2) * g * time^2
h = 10.5 m/s * sin(40 degrees) * [19.6 m / (10.5 m/s * cos(40 degrees))] + (1/2) * 9.8 m/s^2 * [19.6 m / (10.5 m/s * cos(40 degrees))]^2
Now you can calculate this equation to find the height of the building.