A brick is released with no initial speed from the roof of a building and strikes grouond in 2.50s, encountering air drag. How tall, in meters, is the building?
How fast is the brick moving just before it reaches the ground?
Sketch a graph of this falling brick acceleration, velocity and vertical position of time.
To find the height of the building, we can use the equation for the vertical position of an object in free fall:
y = (1/2)gt^2
Where:
y = height of the building
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of fall (2.50s)
Plugging in the values, we get:
y = (1/2)(9.8 m/s^2)(2.50s)^2
= (1/2)(9.8 m/s^2)(6.25s^2)
= 30.625 m
Therefore, the height of the building is 30.625 meters.
To find the velocity of the brick just before it reaches the ground, we can use the equation for the final velocity of an object in free fall:
v = gt
Plugging in the values, we get:
v = (9.8 m/s^2)(2.50s)
= 24.5 m/s
Therefore, the brick is moving at a speed of 24.5 m/s just before it reaches the ground.
To sketch a graph of the acceleration, velocity, and vertical position of the brick over time, we can plot the three quantities on the y-axis and time on the x-axis.
The graph of acceleration would be a constant line at -9.8 m/s^2, representing the acceleration due to gravity.
The graph of velocity would start at 0 m/s and increase linearly until it reaches 24.5 m/s at 2.50s, after which it would remain constant at 24.5 m/s.
The graph of vertical position would start at 0m and increase quadratically with time until it reaches a height of 30.625 m at 2.50s.
To find the height of the building, we can use the following kinematic equation:
h = v₀t + (1/2)gt²
Where:
h = height of the building
v₀ = initial velocity of the brick (which is 0 m/s because it is released with no initial speed)
t = time taken to reach the ground (2.50 s in this case)
g = acceleration due to gravity (approximately 9.8 m/s²)
Plugging in the values:
h = 0 * 2.50 + (1/2) * 9.8 * (2.50)²
h = (1/2) * 9.8 * 6.25
h = 30.625 meters
Therefore, the height of the building is approximately 30.625 meters.
To find the speed of the brick just before it reaches the ground, we can use the following equation:
v = v₀ + gt
Where:
v = final velocity of the brick just before it reaches the ground
v₀ = initial velocity of the brick (which is 0 m/s because it is released with no initial speed)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time taken to reach the ground (2.50 s in this case)
Plugging in the values:
v = 0 + 9.8 * 2.50
v = 24.5 m/s
Therefore, the speed of the brick just before it reaches the ground is 24.5 m/s.
Now, let's sketch the graph of the falling brick's acceleration, velocity, and vertical position with respect to time. Here are the characteristics of each graph:
- Acceleration (a): Constant acceleration due to gravity (approximately 9.8 m/s²), pointing downwards.
- Velocity (v): Starts from 0 m/s, increases linearly with time at a rate of 9.8 m/s², pointing downwards.
- Vertical position (h): Starts from the height of the building, decreases parabolically with time due to gravity.
Please note that I'm unable to sketch the graph visually as a text-based AI, but I hope the descriptions provided help visualize how the graphs would look.
To find the height of the building, we can use the equation for the distance of a falling object:
d = (1/2)gt^2
where d represents the distance, g is the acceleration due to gravity, and t represents the time.
Since the brick is released with no initial speed, its initial velocity is 0. We can use this information to solve for the height (d) of the building.
Given that the brick strikes the ground in 2.50 seconds, we can substitute the values into the equation:
2.50 = (1/2)g(2.50)^2
Simplifying the equation, we get:
2.50 = 1.25g(2.50)^2
Dividing both sides by 1.25(2.50)^2:
2.50 / (1.25 × 2.50^2) = g
Solving for g, we find:
g = 2.50 / (1.25 × 2.50^2)
g ≈ 0.80 m/s^2
Now, we can use the value of g to find the height of the building:
d = (1/2)g(2.50)^2
d = (0.5)(0.80)(2.50^2)
d ≈ 3.13 meters
Therefore, the height of the building is approximately 3.13 meters.
To find the velocity of the brick just before it reaches the ground, we can use the equation for velocity:
v = gt
Given that g = 0.80 m/s^2 and t = 2.50 s, we substitute these values into the equation:
v = 0.80 × 2.50
v ≈ 2.00 m/s
Therefore, the brick is moving at approximately 2.00 meters per second just before it reaches the ground.
To sketch the graph of the falling brick's acceleration, velocity, and vertical position over time, we will need to plot these values on a graph.
The acceleration (a) is constant and equal to -9.8 m/s^2 (negative because it is in the opposite direction to the positive y-axis). The velocity (v) starts from 0 m/s and increases linearly over time, while the vertical position (s) follows a quadratic curve.
The graph will have time (t) on the x-axis and the values of acceleration, velocity, and position on the y-axis. The units for acceleration will be m/s^2, for velocity will be m/s, and for position will be meters.
The sketch will show a linear increase in velocity with time and a curved