a brick is dropped (0 initial speed) from the top of a building. the brick strikes the ground in 2.5 seconds. you may ignore air resistant so the brick is in free fall. A)how tall in meters is the building? B) what is the magnitude of the brick's velocity just before it reaches the ground?
A)
Solve H = (1/2) g T^2 for height, H.
T is the time to fall.
g = 9.8 m/s^2
B)
Solve M g H = (1/2) M V^2, for V.
Note that the mass M cancels out.
Therefore V = sqrt(2gH)
h= 1/2 gT^2
= 1/2 (9.8)(2.5)^2
= 30.63 m/s
To solve this problem, we can use the equations of motion for an object in free fall.
A) To find the height of the building, we need to use the equation:
h = 0.5 * g * t^2
Where:
h is the height of the building
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time it takes for the brick to reach the ground (given as 2.5 seconds)
Plugging in the values, we get:
h = 0.5 * 9.8 * (2.5)^2
h = 0.5 * 9.8 * 6.25
h ≈ 30.625 meters
Therefore, the height of the building is approximately 30.625 meters.
B) To find the magnitude of the brick's velocity just before it reaches the ground, we can use the equation:
v = g * t
Where:
v is the velocity of the brick just before it hits the ground
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time it takes for the brick to reach the ground (given as 2.5 seconds)
Plugging in the values, we get:
v = 9.8 * 2.5
v = 24.5 m/s
Therefore, the magnitude of the brick's velocity just before it reaches the ground is 24.5 m/s.