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)
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?
2 answers
h= 1/2 gT^2
= 1/2 (9.8)(2.5)^2
= 30.63 m/s
= 1/2 (9.8)(2.5)^2
= 30.63 m/s