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?

Question

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.

Answer 1

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.